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MYP 3 · Science

Motion and Forces

100 questions across 5 sub-topics

Use the Sub-Topic filter above to focus on one.

Speed, Distance and Time Distance-Time Graphs Acceleration and Speed-Time Graphs Forces and Newton's Laws (qualitative) Friction, Air Resistance and Terminal Velocity

Speed, Distance and Time 20 questions

QUESTION 1 2 marks Criterion A
Easy

A cyclist covers $150\,\text{m}$ in $30\,\text{s}$ at a constant speed. State the formula for speed, then calculate the cyclist's speed, giving the correct unit.

Show complete worked solution

Step 1 — State the formula:

$$ \text{speed} = \frac{\text{distance}}{\text{time}} = \frac{\Delta s}{\Delta t} $$

Step 2 — Substitute the values:

$$ v = \frac{150}{30} $$

Answer: $v = 5\,\text{m/s}$

QUESTION 2 2 marks Criterion A
Easy

Rearrange the speed formula to make distance the subject. Then calculate the distance travelled by a car moving at $12\,\text{m/s}$ for $15\,\text{s}$.

Show complete worked solution

Rearranged formula: $$ \text{distance} = \text{speed} \times \text{time}, \qquad s = vt $$

Substitute: $$ s = 12 \times 15 $$

Answer: $s = 180\,\text{m}$

QUESTION 3 2 marks Criterion A
Easy

A train travels at an average speed of $80\,\text{km/h}$. Calculate the time taken for the train to travel $200\,\text{km}$. Give your answer in hours, and also state it in minutes.

Show complete worked solution

Rearranged formula: $$ t = \frac{s}{v} = \frac{200}{80} $$

Answer: $t = 2.5\,\text{h}$, which is $2.5 \times 60 = 150\,\text{minutes}$.

QUESTION 4 4 marks Criterion A
Medium

Speeds are often converted between $\text{m/s}$ and $\text{km/h}$.

a. Convert $18\,\text{m/s}$ to $\text{km/h}$.
[2]
b. Convert $54\,\text{km/h}$ to $\text{m/s}$.
[2]
Show complete worked solution
(a)
$$ 18 \times 3.6 = 64.8 $$

Answer: $64.8\,\text{km/h}$

(b)
$$ 54 \div 3.6 = 15 $$

Answer: $15\,\text{m/s}$

QUESTION 5 4 marks Criterion A
Medium

Speed and velocity are related but different quantities.

a. State the difference between speed and velocity.
[2]
b. A car travels once around a circular track of length $400\,\text{m}$ in $80\,\text{s}$, returning exactly to its starting point. State its average speed and its average velocity for the full lap, explaining any difference.
[2]
Show complete worked solution
(a)
Speed is a scalar quantity — it has size (magnitude) only. Velocity is a vector quantity — it has both size and a direction.
(b)
Average speed $= \dfrac{400}{80} = 5\,\text{m/s}$ (total distance travelled). Average velocity $= \dfrac{\text{displacement}}{\text{time}} = \dfrac{0}{80} = 0\,\text{m/s}$, because the car finishes at the same position it started — its overall displacement is zero, even though it travelled $400\,\text{m}$.
QUESTION 6 4 marks Criterion A
Medium

A car's speedometer reads $60\,\text{km/h}$ at one moment during a journey, but the car's average speed for the whole trip works out at only $45\,\text{km/h}$.

a. Explain the difference between these two "speeds".
[2]
b. Explain why average speed alone does not tell you whether the car ever exceeded a speed limit during the journey.
[2]
Show complete worked solution
(a)
The speedometer shows the car's instantaneous speed — its speed at that one exact moment. The $45\,\text{km/h}$ is the average speed for the whole journey, found from total distance $\div$ total time, which blends together faster sections, slower sections, and any stops.
(b)
Average speed only gives one overall value for the whole trip. The car could have travelled well above the speed limit for part of the journey and much more slowly (or been stopped, e.g. at traffic lights) for the rest — these faster and slower periods would average out to $45\,\text{km/h}$ without the average revealing that a limit was briefly exceeded.
QUESTION 7 6 marks Criterion A
Hard

A delivery van travels $18\,\text{km}$ in $15\,\text{minutes}$ on a highway, then travels a further $2400\,\text{m}$ in $8\,\text{minutes}$ through town.

a. Calculate the van's speed for the highway section, in $\text{km/h}$.
[2]
b. Calculate the van's speed for the town section, in $\text{km/h}$.
[2]
c. Calculate the van's average speed for the whole journey, in $\text{km/h}$.
[2]
Show complete worked solution
(a)
$$ t = \frac{15}{60} = 0.25\,\text{h} \qquad v = \frac{18}{0.25} = 72\,\text{km/h} $$
(b)
$$ 2400\,\text{m} = 2.4\,\text{km}, \qquad t = \frac{8}{60} = 0.1\overline{3}\,\text{h} $$$$ v = \frac{2.4}{0.1\overline{3}} = 18\,\text{km/h} $$
(c)
Total distance $= 18 + 2.4 = 20.4\,\text{km}$. Total time $= 15 + 8 = 23\,\text{min} = \dfrac{23}{60}\,\text{h}$.$$ v_{avg} = \frac{20.4}{23/60} = 20.4 \times \frac{60}{23} = 53.2\,\text{km/h} \ (\text{3 s.f.}) $$
QUESTION 8 7 marks Criterion B
Medium

A student wants to find out whether a toy car travels at a different average speed on a wooden floor compared to a carpeted floor.

a. State the independent and dependent variables.
[2]
b. State two variables that should be controlled, and explain why for one of them.
[2]
c. Describe a method, including equipment, to measure and calculate the average speed of the car on each surface.
[3]
Show complete worked solution
(a)
Independent variable: type of floor surface (wood or carpet). Dependent variable: average speed of the toy car (calculated from measured distance and time).
(b)
Control: the toy car used, and the force/push used to launch it each time (e.g. always released from a fixed ramp height). Why control the launch force: if the car were given a different push each trial, any difference in speed could be caused by the push, not the floor surface, making the test unfair.
(c)
  1. Mark a start line and a finish line $2\,\text{m}$ apart on each floor.
  2. Use a fixed ramp to release the car from the same height each time, so the launch force is identical.
  3. Start a stopwatch as the car crosses the start line and stop it as it crosses the finish line.
  4. Repeat $3$ times on each surface and calculate a mean time.
  5. Calculate average speed for each surface using $v = \dfrac{2\,\text{m}}{\text{mean time}}$, and compare the two values.
QUESTION 9 6 marks Criterion B
Medium

A student wants to investigate whether the mass of a trolley affects its speed at the bottom of a fixed ramp.

a. Identify the independent variable, the dependent variable, and one variable that should be controlled.
[3]
b. A student says: "Using a heavier trolley will always give an unreliable result." Evaluate this statement, and suggest what would actually make the results more reliable, regardless of trolley mass.
[3]
Show complete worked solution
(a)
Independent variable: mass of the trolley (e.g. by adding $100\,\text{g}$ masses). Dependent variable: speed of the trolley at the bottom of the ramp (calculated from distance and time, or a light gate). Controlled variable: the angle (steepness) of the ramp and the release point — kept the same in every trial so any change in speed is caused only by the change in mass.
(b)
This statement is not correct — mass by itself does not make results unreliable. Reliability instead depends on how consistently the experiment is carried out. To make results reliable regardless of mass: release the trolley from exactly the same starting point using a repeatable method (e.g. a barrier that is lifted, rather than a hand push), use a light gate or motion sensor to remove reaction-time error, and repeat each mass at least $3$ times and calculate a mean.
QUESTION 10 8 marks Criterion B
Hard

A student measures a sprinter's average speed over $100\,\text{m}$ using a handheld stopwatch, starting the watch as the sprinter leaves the blocks and stopping it as they cross the finish line.

a. Identify the main source of random error in this method, and explain how it would affect the calculated speed.
[3]
b. Suggest one specific improvement to the experimental method that would reduce this error, and explain how it works.
[3]
c. Even with your improvement in (b), explain why repeating the sprint several times and calculating a mean speed is still good practice.
[2]
Show complete worked solution
(a)
The main source of random error is human reaction time (typically $\pm0.2$–$0.3\,\text{s}$) in starting and stopping the stopwatch. This delay adds extra time onto the true race time on some trials more than others, so the calculated time is slightly too long and the calculated speed too low, and the exact size of this error varies unpredictably between trials.
(b)
Use a light-gate (photogate) system at the start and finish lines, connected to an electronic timer, which starts and stops automatically the instant the sprinter's body breaks each light beam. This removes human reaction time from the timing completely, since no person needs to press a button at the exact right moment.
(c)
Even with perfect timing equipment, a sprinter's performance naturally varies slightly from run to run (e.g. due to small differences in effort or technique), so repeating the sprint and averaging gives a value that is more representative of the sprinter's typical speed, rather than relying on a single run that might happen to be unusually fast or slow.
QUESTION 11 3 marks Criterion C
Easy
RunnerDistance (m)Time (s)
A10012.5
C10013.2
a. Calculate the speed of Runner A.
[1]
b. Calculate the speed of Runner C.
[1]
c. Using your answers, state which runner is faster.
[1]
Show complete worked solution
(a)
$$ v = \frac{100}{12.5} = 8\,\text{m/s} $$
(b)
$$ v = \frac{100}{13.2} = 7.58\,\text{m/s} \ (\text{3 s.f.}) $$
(c)
Runner A is faster ($8\,\text{m/s}$ compared to Runner C's $7.58\,\text{m/s}$), consistent with Runner A completing the same $100\,\text{m}$ in less time.
QUESTION 12 4 marks Criterion C
Medium
Trial1234
Time for 5 m (s)1.201.301.252.10
a. Identify the anomalous reading in this table.
[1]
b. Explain your reasoning.
[1]
c. Calculate the mean speed using only the three consistent readings, for a distance of $5\,\text{m}$.
[2]
Show complete worked solution
(a)
Trial $4$ ($1.20\,\text{s}$... rather, the reading of $2.10\,\text{s}$) is anomalous.
(b)
Trials $1$–$3$ are all close together ($1.20$–$1.30\,\text{s}$), but Trial $4$'s time of $2.10\,\text{s}$ is far higher than the other three — it breaks the otherwise consistent pattern, suggesting a mistake such as a late stopwatch start or the ball being released late.
(c)
Mean time $= \dfrac{1.20+1.30+1.25}{3} = \dfrac{3.75}{3} = 1.25\,\text{s}$.$$ v = \frac{5}{1.25} = 4\,\text{m/s} $$
QUESTION 13 5 marks Criterion C
Medium
Distance (km)0–55–1010–1515–20
Time (min)22212423

A marathon runner's split times were recorded every $5\,\text{km}$.

a. Calculate the runner's speed, in $\text{km/h}$, during the $10$–$15\,\text{km}$ segment.
[2]
b. Calculate the runner's overall average speed for the full $20\,\text{km}$, in $\text{km/h}$.
[2]
c. Explain whether the runner's speed was constant throughout, using your data.
[1]
Show complete worked solution
(a)
$$ t = \frac{24}{60} = 0.4\,\text{h} \qquad v = \frac{5}{0.4} = 12.5\,\text{km/h} $$
(b)
Total time $= 22+21+24+23 = 90\,\text{min} = 1.5\,\text{h}$.$$ v_{avg} = \frac{20}{1.5} = 13.3\,\text{km/h} \ (\text{3 s.f.}) $$
(c)
No — the split times vary ($21$ to $24\,\text{minutes}$ per $5\,\text{km}$), so the runner's speed changed during the race: fastest in the $5$–$10\,\text{km}$ segment and slowest in the $10$–$15\,\text{km}$ segment.
QUESTION 14 5 marks Criterion C
Medium
Height (cm)10203040
Time to travel 1 m (s)1.411.000.820.71

A toy car is released from different heights on a ramp and timed over a fixed $1\,\text{m}$ distance at the bottom.

a. Calculate the car's speed when released from a height of $20\,\text{cm}$.
[2]
b. Describe the pattern between height and speed shown by the data.
[2]
c. Predict whether doubling the height from $20\,\text{cm}$ to $40\,\text{cm}$ doubles the speed. Justify your answer using the data.
[1]
Show complete worked solution
(a)
$$ v = \frac{1}{1.00} = 1.00\,\text{m/s} $$
(b)
As height increases, speed increases ($0.71\,\text{m/s} \to 1.41\,\text{m/s}$ across the readings), but the increase in speed gets smaller each time: from $10\,\text{cm}\to20\,\text{cm}$, speed rises by about $0.29\,\text{m/s}$; from $30\,\text{cm}\to40\,\text{cm}$, it rises by only about $0.17\,\text{m/s}$. So speed increases with height, but not in direct proportion.
(c)
No. Speed at $20\,\text{cm}$ is $1.00\,\text{m/s}$; doubling it would give $2.00\,\text{m/s}$, but the actual speed at $40\,\text{cm}$ is only $1.41\,\text{m/s}$ — far less than double.
QUESTION 15 4 marks Criterion C
Medium
Trial1234
Blue car speed (m/s)12111310
Red car speed (m/s)9141011

A student concludes: "The blue car is always faster than the red car."

a. Calculate the mean speed of the blue car and the mean speed of the red car.
[2]
b. Evaluate whether the student's conclusion is fully supported by the data.
[2]
Show complete worked solution
(a)
Blue: $\dfrac{12+11+13+10}{4} = \dfrac{46}{4} = 11.5\,\text{m/s}$. Red: $\dfrac{9+14+10+11}{4} = \dfrac{44}{4} = 11.0\,\text{m/s}$.
(b)
On average the blue car is faster ($11.5\,\text{m/s}$ vs $11.0\,\text{m/s}$), so the conclusion is partly supported. However, it is not always true: in Trial $2$, the red car's speed ($14\,\text{m/s}$) is higher than the blue car's ($11\,\text{m/s}$) — so the word "always" is not supported by the data.
QUESTION 16 7 marks Criterion C
Hard
Trial123456
Time (s)24.123.824.523.924.330.2

Six trials measured the time for a cyclist to cover a fixed $200\,\text{m}$ course.

a. Identify the anomalous reading, and explain your reasoning.
[2]
b. Calculate the mean time using the five consistent readings.
[2]
c. Calculate the cyclist's speed using this mean time.
[2]
d. The five consistent readings have a range of only $0.7\,\text{s}$. Explain what this tells you about the reliability of the repeated results.
[1]
Show complete worked solution
(a)
Trial $6$ ($30.2\,\text{s}$) is anomalous — every other trial falls within a narrow range ($23.8$–$24.5\,\text{s}$), while Trial $6$ is around $6\,\text{s}$ higher, well outside this spread.
(b)
$$ \text{mean} = \frac{24.1+23.8+24.5+23.9+24.3}{5} = \frac{120.6}{5} = 24.12\,\text{s} $$
(c)
$$ v = \frac{200}{24.12} = 8.29\,\text{m/s} \ (\text{3 s.f.}) $$
(d)
A small range ($0.7\,\text{s}$ out of about $24\,\text{s}$) shows the five consistent readings are close together, meaning the method is fairly precise/repeatable — supporting the decision to treat Trial $6$ as a one-off error rather than genuine variation.
QUESTION 17 8 marks Criterion C
Hard
Time (s)03691215
Distance (m)021396379105

A cyclist's position was recorded during a training session.

a. Calculate the cyclist's speed for each of the five $3$-second intervals.
[5]
b. A coach claims the cyclist was accelerating steadily throughout the $15\,\text{s}$. Evaluate this claim using your answers to (a).
[3]
Show complete worked solution
(a)

$0$–$3\,\text{s}$: $\dfrac{21-0}{3}=7.00\,\text{m/s}$

$3$–$6\,\text{s}$: $\dfrac{39-21}{3}=6.00\,\text{m/s}$

$6$–$9\,\text{s}$: $\dfrac{63-39}{3}=8.00\,\text{m/s}$

$9$–$12\,\text{s}$: $\dfrac{79-63}{3}=5.33\,\text{m/s}$

$12$–$15\,\text{s}$: $\dfrac{105-79}{3}=8.67\,\text{m/s}$

(b)
The claim is not supported. If the cyclist were accelerating steadily, the speed should increase in every successive interval. Instead, the calculated speeds rise and fall ($7.00 \to 6.00 \to 8.00 \to 5.33 \to 8.67\,\text{m/s}$), showing alternating periods of speeding up and slowing down rather than one continuous, steady increase — more consistent with varying effort (e.g. on undulating terrain, or a variable-pace training set) than with steady acceleration.
QUESTION 18 6 marks Criterion D
Medium

"Average speed cameras" use two cameras a known distance apart on a motorway. Each camera records a vehicle's number plate and the exact time it passes, so a computer can calculate the vehicle's average speed over that whole stretch of road (rather than its speed at just one single point).

On one stretch, two cameras are $3\,\text{km}$ apart and the speed limit is $100\,\text{km/h}$. A car takes $1\,\text{minute}\ 40\,\text{seconds}$ ($100\,\text{s}$) to travel between them.

Calculate the car's average speed over this stretch, and state whether it appears to have broken the speed limit. Then discuss one benefit and one drawback of using average speed cameras (rather than single-point speed cameras) to enforce speed limits.

Show complete worked solution

Calculation: $$ v = \frac{3\,\text{km}}{100/3600\,\text{h}} = 3 \times 36 = 108\,\text{km/h} $$ This is above the $100\,\text{km/h}$ limit, so the car appears to have broken the speed limit.

Benefit: Because average speed is measured continuously over the whole stretch, a driver cannot simply brake sharply at a single camera location and then speed up again immediately afterwards (as can happen with single-point cameras). This encourages drivers to keep to a consistently safer speed along the entire road, including at points between the two cameras where hazards such as junctions or pedestrians may be present.

Drawback: Because only the average is measured, this system cannot detect a dangerous momentary top speed that still results in a compliant average — for example, a brief unsafe overtaking manoeuvre followed by driving well under the limit for the rest of the stretch. An average speed within the limit does not guarantee the driver's speed was safe and steady the whole way.

QUESTION 19 6 marks Criterion D
Medium

Many cities now require e-scooter companies to fit GPS-based speed limiters that automatically reduce a scooter's motor speed to $10\,\text{km/h}$ in pedestrian-heavy areas, calculated from the scooter's changing GPS position over time.

Discuss one benefit and one drawback of this technology.

Show complete worked solution

Benefit: Automatically reducing speed in crowded areas lowers both the risk and the severity of collisions with pedestrians, since a scooter moving much more slowly gives both the rider and pedestrians more time to react, and a shorter distance is needed to stop — without relying on riders remembering or choosing to slow down themselves.

Drawback: GPS position readings can be inaccurate by several metres, especially between tall buildings, so the speed limiter might engage a little too early or too late at a genuine boundary. Riders may feel unfairly restricted if it engages incorrectly, or pedestrians could be put at risk if it fails to engage in time; continuously tracking the location of individual scooters also raises questions about how that movement data is stored and who can access it.

QUESTION 20 7 marks Criterion D
Hard

A country is considering raising its motorway speed limit from $100\,\text{km/h}$ to $120\,\text{km/h}$. Government data suggests this would reduce journey times, but data from other countries that have raised motorway limits show a rise in serious accident rates afterwards.

Calculate the time saved on a single $200\,\text{km}$ journey if a driver could travel the whole way at $120\,\text{km/h}$ instead of $100\,\text{km/h}$. Then evaluate whether this potential time saving justifies the change, discussing both a benefit and a drawback, and referring to the data given.

Show complete worked solution

Calculation: Time at $100\,\text{km/h}$: $\dfrac{200}{100}=2\,\text{h}=120\,\text{min}$. Time at $120\,\text{km/h}$: $\dfrac{200}{120}=1.667\,\text{h}\approx100\,\text{min}$. Time saved $\approx 20\,\text{minutes}$ per $200\,\text{km}$ trip.

Benefit: A $20$-minute saving on one long trip seems modest, but across many drivers making similar journeys every day, the accumulated time saved is significant — supporting business productivity, reducing driver fatigue on very long trips, and potentially helping emergency vehicles reach incidents faster.

Drawback: The historical data shows higher motorway speed limits are associated with more serious accidents — consistent with higher speeds increasing both stopping distances and the force involved in any collision that does occur, making crashes at $120\,\text{km/h}$ more likely to be fatal than at $100\,\text{km/h}$. Higher speeds also increase air resistance sharply, increasing fuel consumption and CO2 emissions per trip even as journey time falls.

Evaluation: A relatively modest average time saving of around $20$ minutes on a long trip needs to be weighed carefully against a real, well-documented increase in accident risk and severity — a reasonable case for caution rather than a clear-cut decision either way.

Distance-Time Graphs 20 questions

QUESTION 1 2 marks Criterion A
Easy
0246810120163248648096 Time (s) Distance (m)

The graph shows the distance travelled by a cyclist over time.

a. Describe the cyclist's motion between $t=8\,\text{s}$ and $t=12\,\text{s}$.
[1]
b. What does the horizontal section of a distance–time graph always represent?
[1]
Show complete worked solution
(a)
Between $t=8\,\text{s}$ and $t=12\,\text{s}$ the line is horizontal (flat), so distance is not changing while time passes. The cyclist is stationary (at rest) during this interval.
(b)
A horizontal section always represents an object at rest — zero speed, since distance is constant while time increases, giving a gradient of $0$.
QUESTION 2 2 marks Criterion A
Easy

A car travels $150\,\text{m}$ in $10\,\text{s}$ at a constant speed. Calculate the car's speed. State the formula you use, and give your answer with the correct unit.

Show complete worked solution

Step 1 — State the formula:

$$ \text{speed} = \frac{\text{distance}}{\text{time}} = \frac{\Delta s}{\Delta t} $$

Step 2 — Substitute the values:

$$ v = \frac{150}{10} $$

Step 3 — Calculate:

$$ v = 15 $$

Answer: $v = 15\,\text{m/s}$

QUESTION 3 2 marks Criterion A
Easy
Time (s)02468
Distance (m)010203040

Calculate the runner's speed between $t=2\,\text{s}$ and $t=6\,\text{s}$.

Show complete worked solution

Step 1 — Find the change in distance and change in time:

$$ \Delta s = 30 - 10 = 20\,\text{m} \qquad \Delta t = 6 - 2 = 4\,\text{s} $$

Step 2 — Apply the speed formula:

$$ v = \frac{\Delta s}{\Delta t} = \frac{20}{4} $$

Answer: $v = 5\,\text{m/s}$

QUESTION 4 3 marks Criterion A
Medium
0246810120163248648096 Time (s) Distance (m) QP

The graph shows the journeys of two hikers, P and Q, who set off from the same point at the same time.

a. Which hiker, P or Q, is travelling faster? Explain how the graph shows this.
[2]
b. Both lines pass through the origin. What does this tell you about the start of the hike?
[1]
Show complete worked solution
(a)
Hiker Q is travelling faster. On a distance–time graph, speed is shown by the gradient (steepness) of the line — Q's line is steeper than P's, meaning Q covers more distance in the same time.
(b)
Both lines starting at the origin $(0,0)$ shows that both hikers started at the same position at the same time — neither had a head start.
QUESTION 5 4 marks Criterion A
Medium
Time (s)05101520
Distance (m)025505050

This table shows a delivery drone's motion.

a. During which time interval is the drone moving fastest?
[1]
b. Calculate the drone's speed during $0$–$5\,\text{s}$.
[2]
c. What is happening to the drone between $t=10\,\text{s}$ and $t=20\,\text{s}$?
[1]
Show complete worked solution
(a)
Comparing the two moving intervals: $0$–$5\,\text{s}$ covers $25\,\text{m}$, and $5$–$10\,\text{s}$ also covers $25\,\text{m}$ — both are equal, so the drone moves at one constant speed for the whole first $10\,\text{s}$.
(b)
$$ v = \frac{\Delta s}{\Delta t} = \frac{25 - 0}{5 - 0} = 5\,\text{m/s} $$
(c)
Distance stays constant at $50\,\text{m}$ (gradient $=0$), so the drone is stationary / hovering.
QUESTION 6 4 marks Criterion A
Medium

A cheetah runs $120\,\text{m}$ in $4\,\text{s}$.

a. Calculate its speed in $\text{m/s}$.
[2]
b. Convert this speed to $\text{km/h}$.
[2]
Show complete worked solution
(a)
$$ v = \frac{\Delta s}{\Delta t} = \frac{120}{4} = 30\,\text{m/s} $$
(b)
$30\,\text{m/s}$ converts to $\text{km/h}$ using the shortcut multiplier $\times 3.6$:$$ 30 \times 3.6 = 108 $$Answer: $108\,\text{km/h}$
QUESTION 7 6 marks Criterion A
Hard
Time (min)0510152025
Distance (km)044101010

This table records a bus journey.

a. Describe fully what happens during each of the three stages of the journey.
[3]
b. Calculate the bus's average speed, in $\text{km/h}$, for the whole $25$-minute journey (including the stop).
[3]
Show complete worked solution
(a)

Stage 1 (0–5 min): distance rises from $0$ to $4\,\text{km}$ — the bus moves at a constant speed of $\frac{4}{5} = 0.8\,\text{km/min}$.

Stage 2 (5–10 min): distance stays at $4\,\text{km}$ — the bus is stopped (e.g. at a bus stop).

Stage 3 (10–25 min): distance rises from $4$ to $10\,\text{km}$ — the bus moves again at $\frac{10-4}{25-10} = 0.4\,\text{km/min}$, slower than Stage 1 (heavier traffic, perhaps).

(b)

$$ \text{total distance} = 10\,\text{km}, \qquad \text{total time} = \frac{25}{60}\,\text{h} $$

$$ v_{avg} = \frac{10}{25/60} = 10 \times \frac{60}{25} = 24 $$

Answer: $24\,\text{km/h}$

QUESTION 8 7 marks Criterion B
Medium

You want to investigate your own walking speed by producing a distance–time graph from real measurements.

a. State the independent and dependent variables.
[2]
b. State two variables you would need to control, and explain why for one of them.
[2]
c. Describe a method, including the equipment you would use, to collect enough data to plot a distance–time graph.
[3]
Show complete worked solution
(a)
Independent variable: time (measured at set intervals). Dependent variable: distance travelled (measured at each of those times).
(b)
Control the walking surface (e.g. always flat pavement, not slope) and the person doing the walking. Why control the surface: an uneven or sloped surface would change effort and speed for reasons unrelated to normal walking pace, making the test unfair.
(c)
  1. Mark a straight path and measure $20\,\text{m}$ using a trundle wheel, with a marker every $2\,\text{m}$.
  2. Start a stopwatch at $t=0$ as the walker passes the first marker.
  3. Record the time each time the walker passes a $2\,\text{m}$ marker, until $20\,\text{m}$ is reached.
  4. Repeat the walk $3$ times and take a mean time for each distance, to reduce reaction-time error.
  5. Plot distance (y-axis) against mean time (x-axis) to produce the distance–time graph.
  6. QUESTION 9 6 marks Criterion B
    Medium

    A student's hypothesis is: "A ball will roll down a ramp faster as the angle of the ramp increases."

    a. Identify the independent, dependent, and one controlled variable for an investigation to test this.
    [3]
    b. Explain how distance–time data collected for each angle could be used to compare speeds and test the hypothesis.
    [3]
    Show complete worked solution
    (a)
    Independent: angle of the ramp (e.g. $10°,20°,30°,40°$). Dependent: time taken to roll a fixed distance. Controlled: the ball used (same mass and size each time) — a different ball could roll differently regardless of angle.
    (b)
    For each angle, distance and time data (e.g. markers every $0.2\,\text{m}$, timed with a stopwatch or light gates) would be plotted as a distance–time graph. The gradient of each graph gives the ball's speed at that angle. If the hypothesis is correct, steeper angles should give a greater gradient — comparing gradients across all tested angles tests the hypothesis directly.
    QUESTION 10 8 marks Criterion B
    Hard
    Trial12345
    Time (s)1.021.310.981.451.05

    A student measured the time for a trolley to travel a fixed $1\,\text{m}$ down a ramp, repeating the trial $5$ times by hand-timing with a stopwatch.

    a. Identify the main source of error in this method that would explain the spread in these results.
    [2]
    b. Suggest an improvement to the method that would reduce this error, and explain how it works.
    [3]
    c. Explain how the improved method would produce a more reliable distance–time graph.
    [3]
    Show complete worked solution
    (a)
    The main source of error is human reaction time when starting/stopping the stopwatch — this varies trial to trial ($\pm0.2$–$0.3\,\text{s}$), explaining why trials 2 and 4 are noticeably higher.
    (b)
    Replace hand-timing with light gates connected to a data logger/timer at the start and end of the $1\,\text{m}$ distance. The trolley breaking the light beam automatically starts/stops the timer, removing human reaction time entirely.
    (c)
    With reaction-time error removed, repeated trials at the same angle would give very similar times, so data points would sit much closer to a single line rather than being scattered — the gradient calculated would be a far more accurate estimate of the trolley's true speed.
    QUESTION 11 3 marks Criterion C
    Easy
    Time (s)012345
    Distance (m)03691215
    a. Would plotting this data give a straight line or a curve? Explain your reasoning using the numbers.
    [1]
    b. Calculate the speed shown by this data.
    [2]
    Show complete worked solution
    (a)
    A straight line. Each second, distance increases by exactly $3\,\text{m}$ ($0\to3\to6\to9\to12\to15$), so the gradient is constant throughout — the definition of a straight-line (constant-speed) graph.
    (b)
    $$ v = \frac{\Delta s}{\Delta t} = \frac{15 - 0}{5 - 0} = 3\,\text{m/s} $$
    QUESTION 12 4 marks Criterion C
    Medium
    Reading123456
    Time (s)0246810
    Distance (m)048191620
    a. Identify the anomalous reading in this table.
    [1]
    b. Explain how you identified it.
    [2]
    c. State what you should do with this reading before calculating an average speed.
    [1]
    Show complete worked solution
    (a)
    Reading $4$ ($t=6\,\text{s}$, distance $=19\,\text{m}$) is anomalous.
    (b)
    Every other reading increases by exactly $4\,\text{m}$ each time ($0,4,8,\ldots,16,20$). If the pattern had continued, reading $4$ should have been $12\,\text{m}$, not $19\,\text{m}$ — it breaks the otherwise-consistent trend.
    (c)
    Exclude (discount) it from the calculation, and, if possible, repeat that measurement to check whether $19\,\text{m}$ was a recording error, before using the remaining consistent data.
    QUESTION 13 7 marks Criterion C
    Medium
    0246810120163248648096 Time (s) Distance (m)

    The graph shows a cyclist's journey in two stages: a steep section up to $t=4\,\text{s}$, then a shallower section from $t=4\,\text{s}$ to $t=10\,\text{s}$.

    a. Calculate the speed during the first stage ($0$–$4\,\text{s}$).
    [2]
    b. Calculate the speed during the second stage ($4$–$10\,\text{s}$).
    [2]
    c. Calculate the average speed for the whole $10\,\text{s}$ journey, and explain why it differs from your answers to (a) and (b).
    [3]
    Show complete worked solution
    (a)
    $$ v_1 = \frac{60-0}{4-0} = 15\,\text{m/s} $$
    (b)
    $$ v_2 = \frac{90-60}{10-4} = \frac{30}{6} = 5\,\text{m/s} $$
    (c)
    $$ v_{avg} = \frac{90-0}{10-0} = 9\,\text{m/s} $$ This is different from both stage speeds because it is not a simple average of $15$ and $5$ (which would be $10$) — it is weighted by how the whole distance was actually covered over the whole time, blending the faster first stage and slower second stage into one overall rate.
    QUESTION 14 5 marks Criterion C
    Medium
    Time (s)02468
    Distance (m)06121218
    a. Calculate the speed for each $2$-second interval in this table.
    [3]
    b. Describe what the corresponding distance–time graph would look like, referring to your answers in (a).
    [2]
    Show complete worked solution
    (a)

    $0$–$2\,\text{s}$: $v=\dfrac{6-0}{2}=3\,\text{m/s}$

    $2$–$4\,\text{s}$: $v=\dfrac{12-6}{2}=3\,\text{m/s}$

    $4$–$6\,\text{s}$: $v=\dfrac{12-12}{2}=0\,\text{m/s}$

    $6$–$8\,\text{s}$: $v=\dfrac{18-12}{2}=3\,\text{m/s}$

    (b)
    A straight line rising with constant gradient ($3\,\text{m/s}$) from $t=0$ to $t=4\,\text{s}$, then a horizontal section from $t=4$ to $t=6\,\text{s}$ (stopped), then the line rises again with the same gradient from $t=6$ to $t=8\,\text{s}$.
    QUESTION 15 5 marks Criterion C
    Medium
    Time (min)0102030
    Distance walked so far (m)050010001500

    A student walks from home to a park (a straight path, $750\,\text{m}$ away), then walks back home along the same path, at a constant speed the whole time. The table shows the total distance the student's feet have covered, not their distance from home.

    a. At what time does the student reach the park and turn around?
    [1]
    b. Explain why the student's distance from home at $t=30\,\text{min}$ is $0\,\text{m}$, even though the table shows $1500\,\text{m}$ walked in total.
    [2]
    c. Calculate the student's walking speed.
    [2]
    Show complete worked solution
    (a)
    $t=15\,\text{min}$ — halfway through the trip, since $750\,\text{m}$ is halfway between the start and the $1000\,\text{m}$-total point at $t=20\,\text{min}$.
    (b)
    The table records total distance travelled (like an odometer) — it keeps increasing the whole walk, there and back. Distance from home is different: it rises to $750\,\text{m}$ on the way out, then falls back to $0\,\text{m}$ on the way back, since the student ends up where they started.
    (c)
    $$ v = \frac{1500}{30} = 50\,\text{m/min} $$
    QUESTION 16 8 marks Criterion C
    Hard
    Trial123456
    Time (s)1.02.12.94.24.86.1
    Distance (m)24681012
    a. Calculate the speed between each consecutive pair of readings (5 intervals).
    [5]
    b. Using your answers, evaluate whether the motion was constant speed. Support your conclusion using the size of the variation you calculated.
    [3]
    Show complete worked solution
    (a)

    1?2: $\frac{4-2}{2.1-1.0}=1.82\,\text{m/s}$

    2?3: $\frac{6-4}{2.9-2.1}=2.50\,\text{m/s}$

    3?4: $\frac{8-6}{4.2-2.9}=1.54\,\text{m/s}$

    4?5: $\frac{10-8}{4.8-4.2}=3.33\,\text{m/s}$

    5?6: $\frac{12-10}{6.1-4.8}=1.54\,\text{m/s}$

    (b)
    The calculated speeds vary widely, from about $1.54$ to $3.33\,\text{m/s}$ — more than double between smallest and largest. If motion were genuinely constant, these should all be close together. This spread is most likely explained by reaction-time error in hand-timing rather than the speed truly changing five times — the distance readings themselves increase by an equal $2\,\text{m}$ each time, supporting approximately constant speed despite the noisy timing.
    QUESTION 17 7 marks Criterion C
    Hard
    Trial12345
    Time for 10 m (s)2.312.282.402.262.35
    a. Calculate the mean time for this data.
    [2]
    b. Calculate the speed using the mean time.
    [2]
    c. The stopwatch could be read to $\pm0.01\,\text{s}$, but the readings vary by more than this. Explain what this tells you about the main source of uncertainty in this experiment.
    [3]
    Show complete worked solution
    (a)
    $$ \text{mean} = \frac{2.31+2.28+2.40+2.26+2.35}{5} = \frac{11.60}{5} = 2.32\,\text{s} $$
    (b)
    $$ v = \frac{10}{2.32} = 4.31\,\text{m/s} \ (\text{3 s.f.}) $$
    (c)
    Since the readings vary by up to $0.14\,\text{s}$ — far more than the stopwatch's own $\pm0.01\,\text{s}$ precision — the main source of uncertainty is not the instrument, but human reaction time in starting/stopping the stopwatch. This is why repeating and averaging trials is essential.
    QUESTION 18 4 marks Criterion D
    Medium

    A car travelling at $30\,\text{m/s}$ takes about $2\,\text{s}$ to react and brake once a hazard is seen, then a further distance to stop once the brakes are applied.

    Using ideas about distance, time and speed, explain why understanding an object's speed from a distance–time graph is important for setting safe speed limits near schools.

    Show complete worked solution

    The distance a car covers in its "thinking time" (before braking even begins) is speed $\times$ time — a faster car covers a much greater distance in the same reaction time than a slower one. At $30\,\text{m/s}$, in a $2\,\text{s}$ reaction time the car has already travelled $30\times2=60\,\text{m}$ before braking starts, before any braking distance is added. Near a school, where children may step into the road with little warning, a lower speed limit directly reduces both this "thinking distance" and the total stopping distance, giving the driver more time and a shorter distance to stop safely. This is exactly the relationship shown by the gradient of a distance–time graph: a steeper gradient (higher speed) means more distance is covered for every second that passes — which is why speed limits are set lower where sudden hazards are more likely.

    QUESTION 19 6 marks Criterion D
    Medium
    Road typeWithout speed bumpsWith speed bumps
    Average speed (km/h)4822
    Time for 1 km (min)1.252.7

    A local council is deciding whether to install speed bumps on a residential road. The table shows average speed data collected before and after a trial installation nearby.

    Discuss one benefit and one drawback of installing speed bumps, using the data to support your discussion.

    Show complete worked solution

    Benefit: Average speed dropped substantially, from $48\,\text{km/h}$ to $22\,\text{km/h}$, more than halving it. A lower speed reduces both the distance a vehicle travels during a driver's reaction time and its stopping distance once braking, likely reducing the number and severity of accidents on a residential road, especially where pedestrians are present.

    Drawback: The time to travel $1\,\text{km}$ more than doubled, from $1.25$ to $2.7$ minutes. Over many journeys, and for emergency vehicles like ambulances that may need to use the road, this extra time could be significant — delaying urgent responses, as well as increasing journey times, fuel use, and vehicle wear for everyday drivers.

    QUESTION 20 6 marks Criterion D
    Hard

    Modern smartphones can use GPS to record a person's position many times per second, allowing an app to automatically generate a distance–time graph of a run, cycle, or drive, and calculate average and maximum speed.

    Evaluate the impact of this technology, discussing both a benefit and a concern it raises for society.

    Show complete worked solution

    Benefit: This gives athletes, cyclists, and everyday users far more precise and effortless speed and distance data than manual timing methods, removing sources of human error such as reaction time. It supports training, and safety features such as automatically alerting emergency services with a precise location if a cyclist stops moving unexpectedly after a fall.

    Concern: Continuously recording a person's exact location and speed raises privacy concerns — this data reveals detailed patterns about where a person lives, works, and travels, and when. If stored by a company, shared with third parties, or accessed without permission, it could be misused to track someone's movements without their consent — which is why data-protection regulations increasingly require apps to be transparent and obtain clear consent before recording or sharing movement data.

    Acceleration and Speed-Time Graphs 20 questions

    QUESTION 1 2 marks Criterion A
    Easy

    A car speeds up from $0\,\text{m/s}$ to $20\,\text{m/s}$ in $8\,\text{s}$. State the formula for acceleration, then calculate the car's acceleration.

    Show complete worked solution

    Formula: $$ a = \frac{\Delta v}{\Delta t} = \frac{v - u}{t} $$

    Substitute: $$ a = \frac{20 - 0}{8} $$

    Answer: $a = 2.5\,\text{m/s}^2$

    QUESTION 2 2 marks Criterion A
    Easy

    A cyclist decelerates from $12\,\text{m/s}$ to $4\,\text{m/s}$ in $4\,\text{s}$. Calculate the cyclist's acceleration, and state whether this represents speeding up or slowing down.

    Show complete worked solution
    $$ a = \frac{4 - 12}{4} = \frac{-8}{4} = -2\,\text{m/s}^2 $$

    The negative sign shows the cyclist is slowing down (decelerating) — the magnitude of the deceleration is $2\,\text{m/s}^2$.

    QUESTION 3 2 marks Criterion A
    Easy
    0 2 4 6 8 10 0 2 4 6 8 Time (s) Speed (m/s)

    The graph shows the speed of a lift (elevator) over a $10\,\text{s}$ period.

    a. Describe what is happening to the lift's speed during this time.
    [1]
    b. Calculate the lift's acceleration during this time.
    [1]
    Show complete worked solution
    (a)
    The line is horizontal (flat) at $6\,\text{m/s}$, so the lift's speed is constant — not changing.
    (b)
    Since speed does not change ($\Delta v = 0$), $$ a = \frac{\Delta v}{\Delta t} = \frac{0}{10} = 0\,\text{m/s}^2 $$
    QUESTION 4 4 marks Criterion A
    Medium

    A skateboarder starts at $2\,\text{m/s}$ and accelerates at $1.5\,\text{m/s}^2$ for $6\,\text{s}$.

    a. State the formula linking final speed $v$, initial speed $u$, acceleration $a$ and time $t$, explaining what each symbol means.
    [2]
    b. Calculate the skateboarder's final speed.
    [2]
    Show complete worked solution
    (a)
    $$ v = u + at $$ where $v$ = final speed, $u$ = initial (starting) speed, $a$ = acceleration, and $t$ = time taken.
    (b)
    $$ v = u + at = 2 + (1.5 \times 6) = 2 + 9 $$

    Answer: $v = 11\,\text{m/s}$

    QUESTION 5 4 marks Criterion A
    Medium
    0 1 2 3 4 5 6 0 5 10 15 20 Time (s) Speed (m/s)

    The graph shows a trolley's speed as it accelerates uniformly from rest.

    a. Calculate the acceleration shown by the graph.
    [2]
    b. Assuming the acceleration stays constant, use the gradient to calculate the trolley's speed at $t=4\,\text{s}$.
    [2]
    Show complete worked solution
    (a)
    $$ a = \frac{\Delta v}{\Delta t} = \frac{20 - 5}{6 - 0} = \frac{15}{6} $$

    Answer: $a = 2.5\,\text{m/s}^2$

    (b)
    $$ v = u + at = 5 + (2.5 \times 4) = 5 + 10 $$

    Answer: $v = 15\,\text{m/s}$

    QUESTION 6 4 marks Criterion A
    Medium

    Acceleration can be positive, negative, or zero.

    a. State the units of acceleration, and explain what they mean in terms of how speed changes.
    [2]
    b. A skydiver's acceleration is recorded as $-2\,\text{m/s}^2$ at one point during a fall. Explain what the negative sign tells you about the skydiver's motion at that point.
    [2]
    Show complete worked solution
    (a)
    The units are metres per second squared ($\text{m/s}^2$). This means the object's speed (in $\text{m/s}$) changes by that many $\text{m/s}$ every second.
    (b)
    The negative sign shows the skydiver is decelerating — their speed is decreasing over time (for example, shortly after their parachute opens), rather than speeding up.
    QUESTION 7 6 marks Criterion A
    Hard

    A high-speed train accelerates uniformly from $20\,\text{m/s}$ to $65\,\text{m/s}$ over $90\,\text{s}$, then continues at a constant $65\,\text{m/s}$ for a further $300\,\text{s}$.

    a. Calculate the train's acceleration during the first $90\,\text{s}$.
    [2]
    b. State the train's acceleration during the next $300\,\text{s}$.
    [1]
    c. Calculate the train's speed $40\,\text{s}$ after it starts accelerating (i.e. still within the first $90\,\text{s}$), assuming the acceleration is uniform.
    [3]
    Show complete worked solution
    (a)
    $$ a = \frac{65 - 20}{90} = \frac{45}{90} $$

    Answer: $a = 0.5\,\text{m/s}^2$

    (b)
    $a = 0\,\text{m/s}^2$, since the train's speed is constant ($65\,\text{m/s}$) during this time.
    (c)
    $$ v = u + at = 20 + (0.5 \times 40) = 20 + 20 $$

    Answer: $v = 40\,\text{m/s}$

    QUESTION 8 7 marks Criterion B
    Medium

    A student wants to investigate how the angle of a ramp affects the acceleration of a trolley released from rest.

    a. State the independent variable, the dependent variable, and one variable that must be controlled.
    [3]
    b. Describe how two light gates positioned along the ramp could be used to calculate the trolley's acceleration.
    [4]
    Show complete worked solution
    (a)
    Independent variable: angle of the ramp (e.g. $10^\circ, 20^\circ, 30^\circ, 40^\circ$). Dependent variable: acceleration of the trolley. Controlled variable: the trolley used (mass) and the distance between the two light gates — kept the same each trial so results are comparable and any change is due only to the angle.
    (b)
    Place light gate $1$ near the top of the ramp and light gate $2$ a fixed distance further down. Each gate, connected to a datalogger, records the trolley's speed as it passes (using the known length of the card mounted on the trolley and the time it takes to break the beam), and the datalogger also records the time between the two gates. Acceleration is then calculated using $$ a = \frac{v_2 - v_1}{t} $$ where $v_1$ and $v_2$ are the speeds at gate $1$ and gate $2$, and $t$ is the time between the gates. Repeat and average for each angle to improve reliability.
    QUESTION 9 6 marks Criterion B
    Medium

    A student uses a ball dropped from different heights and a stopwatch to try to find the acceleration of a falling object due to gravity.

    a. State the independent and dependent variables for this investigation.
    [2]
    b. Explain one reason why using a stopwatch to time a free-falling object gives unreliable acceleration values, and suggest a more suitable piece of equipment.
    [4]
    Show complete worked solution
    (a)
    Independent variable: height the ball is dropped from. Dependent variable: acceleration (calculated from the height and the time of fall).
    (b)
    Free fall over a typical drop height happens very quickly (often under a second), so human reaction-time error (roughly $\pm0.2$–$0.3\,\text{s}$) makes up a very large fraction of the total time being measured. This makes the recorded time — and so the calculated acceleration — highly inaccurate and inconsistent between repeats. A light-gate system or motion sensor that starts and stops timing automatically as the object passes fixed points removes this reaction-time error and gives a far more precise, repeatable measurement.
    QUESTION 10 8 marks Criterion B
    Hard

    A student uses a motion sensor connected to a computer to record a trolley's speed every $0.1\,\text{s}$ as it accelerates down a ramp. The computer automatically calculates the gradient of the resulting speed-time graph as the trolley's acceleration.

    a. Explain one advantage of using a motion sensor and computer over a student using a stopwatch and metre ruler to find acceleration.
    [3]
    b. The student obtains a slightly different acceleration value on three repeated trials down the same ramp ($1.98$, $2.05$, $1.93\,\text{m/s}^2$). Suggest one experimental reason for this small variation (not equipment fault), and one way to reduce its effect on the final reported result.
    [3]
    c. Explain why placing the motion sensor at the very top of the ramp, right where the trolley starts from rest, could cause a systematic error in the measured acceleration.
    [2]
    Show complete worked solution
    (a)
    A motion sensor takes many measurements automatically and very rapidly (every $0.1\,\text{s}$), removing human reaction-time error entirely and providing enough data points to plot a smooth, detailed speed-time graph. A stopwatch and ruler can typically only give one or two averaged speed values, each affected by significant reaction-time error.
    (b)
    Small real variations can occur trial to trial even with the same equipment — for example, slightly different friction along the track, or the trolley not being released from exactly the same starting position each time. To reduce the effect on the final result, repeat the trial several more times and calculate a mean acceleration, discarding any clear outlier if one appears.
    (c)
    Right at the very start of motion, the trolley's speed is very low, and the earliest readings can be affected by small alignment or start-up detection errors in the sensor. If these unreliable early data points are included when calculating the overall gradient (acceleration), they could consistently shift the calculated value away from the trolley's true, steady acceleration.
    QUESTION 11 3 marks Criterion C
    Easy
    Time (s)0246
    Speed (m/s)061218
    a. Calculate the acceleration between $t=0\,\text{s}$ and $t=4\,\text{s}$.
    [2]
    b. State whether the acceleration is the same across all three $2$-second intervals shown, and briefly justify your answer.
    [1]
    Show complete worked solution
    (a)
    $$ a = \frac{12 - 0}{4 - 0} = 3\,\text{m/s}^2 $$
    (b)
    Yes — the speed increases by exactly $6\,\text{m/s}$ every $2\,\text{s}$ throughout the table, so the acceleration is constant ($3\,\text{m/s}^2$) across the whole $6\,\text{s}$.
    QUESTION 12 5 marks Criterion C
    Medium
    0 2 4 6 8 10 12 0 5 10 15 20 Time (s) Speed (m/s)

    The graph shows the speed of a delivery drone over a $12\,\text{s}$ flight, made of three stages.

    a. Describe the motion of the drone during each of the three stages shown.
    [3]
    b. Calculate the total distance travelled by the drone during the constant-speed stage ($4\,\text{s}$ to $10\,\text{s}$), using the area under the graph.
    [2]
    Show complete worked solution
    (a)
    Stage 1 (0–4 s): the line rises steadily — the drone accelerates from rest up to $20\,\text{m/s}$. Stage 2 (4–10 s): the line is flat at $20\,\text{m/s}$ — the drone travels at a constant speed. Stage 3 (10–12 s): the line falls back to $0$ — the drone decelerates to a stop.
    (b)
    This section is a rectangle: $$ \text{distance} = \text{speed} \times \text{time} = 20 \times (10-4) = 20 \times 6 $$

    Answer: $120\,\text{m}$

    QUESTION 13 5 marks Criterion C
    Medium
    Time (s)012345
    Speed (m/s)02413810
    a. Identify the anomalous reading in this table.
    [1]
    b. Explain your reasoning using the pattern in the rest of the data.
    [2]
    c. Calculate the acceleration shown by the data, using the readings at $t=0\,\text{s}$ and $t=5\,\text{s}$ only (unaffected by the anomaly).
    [2]
    Show complete worked solution
    (a)
    The reading at $t=3\,\text{s}$ ($13\,\text{m/s}$) is anomalous.
    (b)
    Every other value increases by exactly $2\,\text{m/s}$ per second ($0,2,4,\_,8,10$), showing a constant acceleration of $2\,\text{m/s}^2$. If this pattern had continued, the reading at $t=3\,\text{s}$ should have been $6\,\text{m/s}$, not $13\,\text{m/s}$ — this breaks the otherwise consistent trend.
    (c)
    $$ a = \frac{10 - 0}{5 - 0} = 2\,\text{m/s}^2 $$
    QUESTION 14 6 marks Criterion C
    Medium
    0 3 6 9 0 5 10 15 20 25 Time (s) Speed (m/s)

    The graph shows a motorbike's speed over $9\,\text{s}$, made of two stages with different gradients.

    a. Calculate the acceleration during the first stage ($0$–$3\,\text{s}$).
    [2]
    b. Calculate the acceleration during the second stage ($3$–$9\,\text{s}$).
    [2]
    c. Without calculating exact areas, state which stage covers the greater distance, and justify your answer by comparing the two areas under the graph.
    [2]
    Show complete worked solution
    (a)
    $$ a_1 = \frac{15 - 0}{3 - 0} = 5\,\text{m/s}^2 $$
    (b)
    $$ a_2 = \frac{21 - 15}{9 - 3} = \frac{6}{6} = 1\,\text{m/s}^2 $$
    (c)
    Stage $2$ covers the greater distance. Stage $1$'s area is a triangle covering roughly $0.5 \times 3 \times 15 = 22.5\,\text{m}$. Stage $2$'s area is a (larger) trapezium lasting twice as long ($6\,\text{s}$) and reaching higher speeds ($15$–$21\,\text{m/s}$), giving roughly $0.5\times(15+21)\times6 = 108\,\text{m}$ — clearly larger than Stage $1$, even though its acceleration is smaller.
    QUESTION 15 5 marks Criterion C
    Medium
    Time (s)0123456
    Speed (m/s)24201612840

    This table records a car braking to a stop.

    a. Show that this data indicates a constant deceleration, using calculated values.
    [2]
    b. Calculate the total braking distance using the area under a speed-time graph of this data (a triangle from $(0,24)$ to $(6,0)$).
    [3]
    Show complete worked solution
    (a)
    Each second, the speed falls by exactly $4\,\text{m/s}$ ($24\to20\to16\to12\to8\to4\to0$). Since the gradient (rate of change of speed) is the same in every interval, this gives a constant deceleration of $4\,\text{m/s}^2$ throughout.
    (b)
    $$ \text{distance} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 24 $$

    Answer: $72\,\text{m}$

    QUESTION 16 8 marks Criterion C
    Hard
    0 2 4 6 8 10 12 14 0 4 8 12 16 Time (s) Speed (m/s)

    The graph shows a car's speed over a $14\,\text{s}$ period: accelerating, then travelling at a constant speed, then decelerating to a stop.

    a. Calculate the acceleration during $0$–$4\,\text{s}$.
    [2]
    b. Calculate the deceleration during $10$–$14\,\text{s}$.
    [2]
    c. Calculate the total distance travelled over the whole $14\,\text{s}$, using the sum of the areas of the three sections.
    [3]
    d. Calculate the car's average speed for the whole $14\,\text{s}$ journey.
    [1]
    Show complete worked solution
    (a)
    $$ a = \frac{16 - 0}{4} = 4\,\text{m/s}^2 $$
    (b)
    $$ a = \frac{0 - 16}{14 - 10} = \frac{-16}{4} = -4\,\text{m/s}^2, \text{ magnitude } 4\,\text{m/s}^2 $$
    (c)
    Section 1 (triangle): $\frac{1}{2}\times4\times16=32\,\text{m}$. Section 2 (rectangle): $6\times16=96\,\text{m}$. Section 3 (triangle): $\frac{1}{2}\times4\times16=32\,\text{m}$.$$ \text{total} = 32+96+32 = 160\,\text{m} $$
    (d)
    $$ v_{avg} = \frac{160}{14} = 11.4\,\text{m/s} \ (\text{3 s.f.}) $$
    QUESTION 17 7 marks Criterion C
    Hard
    Trial12345
    Time to reach 2 m/s from rest (s)1.020.951.100.981.05
    a. Calculate the mean time.
    [2]
    b. Calculate the acceleration using the mean time (from rest to $2\,\text{m/s}$).
    [2]
    c. The range of the five times is $0.15\,\text{s}$ ($1.10-0.95$). Explain what this suggests about the precision of this method, and suggest one way to improve it.
    [3]
    Show complete worked solution
    (a)
    $$ \text{mean} = \frac{1.02+0.95+1.10+0.98+1.05}{5} = \frac{5.10}{5} = 1.02\,\text{s} $$
    (b)
    $$ a = \frac{2 - 0}{1.02} = 1.96\,\text{m/s}^2 \ (\text{3 s.f.}) $$
    (c)
    A range of $0.15\,\text{s}$ is fairly large compared to the times themselves (about $15\%$ of the mean), suggesting the timing method (likely a hand-operated stopwatch, affected by reaction time) is not very precise or repeatable. Using a light gate or motion sensor to detect automatically when the trolley reaches $2\,\text{m/s}$ would remove reaction-time error and give a smaller range of results — a more precise, consistent method.
    QUESTION 18 6 marks Criterion D
    Medium

    Modern cars are fitted with crumple zones — sections of the car's frame designed to collapse gradually during a collision, increasing the time over which the car (and its occupants) decelerate to a stop, rather than stopping almost instantly.

    Discuss one benefit and one drawback of this design feature, referring to the effect on acceleration.

    Show complete worked solution

    Benefit: For the same change in speed (e.g. stopping from $15\,\text{m/s}$ to $0$), increasing the time taken to stop reduces the size of the deceleration, since $a = \dfrac{\Delta v}{\Delta t}$ — a larger $\Delta t$ gives a smaller $a$. Smaller decelerations mean smaller forces act on the occupants during the crash, significantly reducing the risk of serious injury compared to a rigid car that stops almost instantly.

    Drawback: Crumple zones are designed to be permanently damaged in a collision, so even a moderate crash can be very expensive to repair or can write off the car entirely. Designing and manufacturing more complex crash structures can also increase the overall cost and weight of the vehicle.

    QUESTION 19 6 marks Criterion D
    Medium

    Elite sprinters increasingly train using wearable accelerometers that record their acceleration many times per second during a race, allowing coaches to identify exactly which part of the race (for example, the first $2\,\text{s}$ out of the blocks) needs the most improvement.

    Discuss one benefit and one drawback of this technology.

    Show complete worked solution

    Benefit: Precise acceleration data lets coaches pinpoint exactly where in the race an athlete loses time (for example, a slower-than-average acceleration phase out of the blocks) far more accurately than watching by eye. This allows training to be targeted efficiently, helping athletes improve performance and pursue faster times.

    Drawback: This equipment, and the expertise needed to interpret its data, can be expensive, so athletes and teams with more funding have access to more precise training feedback than others. This could widen the performance gap between well-funded and under-funded athletes or countries, rather than the sport remaining a level playing field.

    QUESTION 20 7 marks Criterion D
    Hard

    Electric vehicles (EVs) can typically accelerate from rest much faster than a similarly priced petrol car, because an electric motor delivers its maximum turning force instantly, whereas a petrol engine needs to build up speed before delivering full power.

    Evaluate the impact of this on society, discussing both a benefit and a concern.

    Show complete worked solution

    Benefit: Greater acceleration is a genuine safety and performance advantage in some situations — for example, accelerating quickly onto a busy motorway from a slip road, or overtaking safely, can reduce the time spent in a dangerous position relative to other traffic. Combined with EVs producing no exhaust emissions, this delivers both a performance and an environmental improvement over petrol cars.

    Concern: Very fast, near-silent acceleration in urban areas increases risk to pedestrians and cyclists, who often rely partly on engine noise to judge how quickly an approaching vehicle is moving — a much higher-than-expected acceleration from a car that sounds like it is barely moving could catch pedestrians off guard, especially children or visually impaired people. High accelerations, especially in heavier EVs (due to battery weight), also increase tyre wear, releasing more microplastic particles into the environment despite there being no exhaust emissions.

    Forces and Newton's Laws (qualitative) 20 questions

    QUESTION 1 2 marks Criterion A
    Easy

    State Newton's First Law of Motion, in your own words.

    Show complete worked solution

    An object will remain at rest, or continue moving at a constant velocity (constant speed in a straight line), unless a resultant (unbalanced) force acts on it.

    QUESTION 2 2 marks Criterion A
    Easy

    Give one example of a contact force and one example of a non-contact force, explaining the difference between the two types.

    Show complete worked solution

    Contact force (e.g. friction, or a push): requires the two objects to be touching for the force to act. Non-contact force (e.g. gravity, or magnetism): acts between two objects even when they are not touching, across a gap.

    QUESTION 3 2 marks Criterion A
    Easy
    Box 4 N 4 N

    The diagram shows two horizontal forces acting on a box resting on the ground.

    a. Calculate the resultant force acting on the box.
    [1]
    b. State whether the forces are balanced or unbalanced, and what this means for the box's motion.
    [1]
    Show complete worked solution
    (a)
    The two forces are equal ($4\,\text{N}$) and opposite, so they cancel out: $$ 4 - 4 = 0\,\text{N} $$
    (b)
    The forces are balanced (resultant $=0\,\text{N}$). By Newton's First Law, if the box was already at rest, it will remain at rest; if it was already moving, it would continue at a constant velocity.
    QUESTION 4 4 marks Criterion A
    Medium

    Newton's Third Law describes pairs of forces.

    a. State Newton's Third Law.
    [2]
    b. A book rests on a table. Identify the reaction-force pair to "the table pushes up on the book", stating its size, direction, and which object it acts on.
    [2]
    Show complete worked solution
    (a)
    For every action force, there is an equal and opposite reaction force.
    (b)
    The book pushes down on the table, with a force equal in size to the table's force on the book, but acting in the opposite direction (downward), and acting on the table (not on the book).
    QUESTION 5 4 marks Criterion A
    Medium

    A resultant force of $12\,\text{N}$ acts on a trolley of mass $4\,\text{kg}$.

    a. Calculate the trolley's acceleration.
    [2]
    b. State what would happen to the acceleration if the same force acted on a trolley with twice the mass, without recalculating a new numeric value.
    [2]
    Show complete worked solution
    (a)
    $$ a = \frac{F}{m} = \frac{12}{4} $$

    Answer: $a = 3\,\text{m/s}^2$

    (b)
    The acceleration would be halved (it would become $1.5\,\text{m/s}^2$). For the same force, acceleration is inversely proportional to mass ($a=F/m$), so doubling the mass halves the acceleration.
    QUESTION 6 4 marks Criterion A
    Medium

    In a tug of war, Team A pulls with a force of $850\,\text{N}$ and Team B pulls with a force of $700\,\text{N}$, in opposite directions.

    a. Calculate the resultant force on the rope.
    [2]
    b. State which team wins initially, and explain why, referring to Newton's First Law.
    [2]
    Show complete worked solution
    (a)
    $$ 850 - 700 = 150\,\text{N, in Team A's direction} $$
    (b)
    Team A wins initially. There is a resultant (unbalanced) force of $150\,\text{N}$ in Team A's direction, so by Newton's First Law, the rope — which was at rest with balanced forces — can no longer stay at rest. It accelerates in the direction of the resultant force, towards Team A.
    QUESTION 7 6 marks Criterion A
    Hard

    A rocket-powered toy car of mass $2\,\text{kg}$ is initially at rest. A resultant force of $5\,\text{N}$ acts on it for $4\,\text{s}$.

    a. Calculate the acceleration produced.
    [2]
    b. Calculate the car's speed after the $4\,\text{s}$.
    [2]
    c. The motor is then adjusted so the same $5\,\text{N}$ resultant force acts on a car of unknown mass, producing an acceleration of $1\,\text{m/s}^2$. Calculate this new mass.
    [2]
    Show complete worked solution
    (a)
    $$ a = \frac{F}{m} = \frac{5}{2} $$

    Answer: $a = 2.5\,\text{m/s}^2$

    (b)
    $$ v = u + at = 0 + (2.5 \times 4) $$

    Answer: $v = 10\,\text{m/s}$

    (c)
    $$ m = \frac{F}{a} = \frac{5}{1} $$

    Answer: $m = 5\,\text{kg}$

    QUESTION 8 6 marks Criterion B
    Medium

    A student wants to investigate how the mass of a trolley affects the acceleration produced by a constant applied force.

    a. State the independent variable, the dependent variable, and one variable that must be controlled.
    [3]
    b. Describe how a constant force could be applied to the trolley for every trial, and how the resulting acceleration could be measured.
    [3]
    Show complete worked solution
    (a)
    Independent variable: mass of the trolley (added masses). Dependent variable: acceleration of the trolley. Controlled variable: the applied (pulling) force — kept the same in every trial, for example by always using the same falling hanging mass on a string over a pulley.
    (b)
    Attach a string over a pulley at the edge of the table to the trolley, with a fixed hanging mass (e.g. $50\,\text{g}$) providing the same pulling force in every trial. Use light gates (or a motion sensor) positioned along the track to record the trolley's speed at two points and the time between them, then calculate $a = \dfrac{\Delta v}{\Delta t}$. Repeat for trolleys loaded with different added masses, keeping the hanging mass the same each time.
    QUESTION 9 6 marks Criterion B
    Medium

    A student wants to investigate the relationship between the size of an applied force and the acceleration it produces, for a trolley of fixed mass.

    a. State the independent and dependent variables, and one variable that must be controlled.
    [3]
    b. Describe how a newton-meter (force meter) could be used to apply and measure a range of different, known forces to the trolley.
    [3]
    Show complete worked solution
    (a)
    Independent variable: size of the applied (pulling) force. Dependent variable: acceleration produced. Controlled variable: the mass of the trolley — kept constant throughout, so any change in acceleration is caused only by the change in force.
    (b)
    Attach a newton-meter to the trolley and pull it along a smooth, level track, reading the newton-meter's scale to keep the applied force at a chosen constant value (e.g. $2\,\text{N}$) for that trial, using light gates or a motion sensor to measure the resulting acceleration. Repeat with different newton-meter readings (e.g. $4\,\text{N}$, $6\,\text{N}$, $8\,\text{N}$), keeping the trolley's mass the same throughout, then compare whether doubling the force doubles the acceleration.
    QUESTION 10 8 marks Criterion B
    Hard

    In an experiment to test the relationship between force and acceleration, a student pulls a trolley across a table using a string over a pulley, with different hanging masses providing the force, while keeping the trolley's own mass constant. The student ignores the friction between the trolley and the table in her calculations.

    a. Explain why ignoring friction is likely to introduce a systematic error into the student's results.
    [3]
    b. Explain whether this error would affect every trial equally, or affect trials with a small hanging force more than trials with a large hanging force, and why.
    [3]
    c. Suggest one way the student could reduce the effect of friction on her results.
    [2]
    Show complete worked solution
    (a)
    In reality, the resultant force accelerating the trolley is smaller than the string's tension, because friction between the trolley and the table acts backwards, opposing the motion. By assuming the pulling force is the resultant force (ignoring friction), the student consistently calculates an accelerating force that is too large for the acceleration actually produced — a systematic error affecting every reading in the same direction.
    (b)
    It would affect trials with a small hanging force more, proportionally. Friction stays roughly the same size regardless of the pulling force, so it makes up a much larger fraction of a small applied force than of a large one — at low forces, a bigger proportion of the applied force is "used up" overcoming friction rather than accelerating the trolley, distorting the results more at low forces.
    (c)
    Use a smooth, low-friction track (e.g. an air track, or a track with rollers), or first measure the amount of friction present and add a small amount of extra hanging mass to compensate for it ("frictional compensation") before starting the actual trials, so the trolley moves at a near-constant speed with no additional pulling force.
    QUESTION 11 3 marks Criterion C
    Easy
    TrialForce left (N)Force right (N)
    1106
    288
    359
    a. Calculate the resultant force for Trial 1, stating its direction.
    [1]
    b. State which trial has balanced forces.
    [1]
    c. Calculate the resultant force for Trial 3, stating its direction.
    [1]
    Show complete worked solution
    (a)
    $$ 10 - 6 = 4\,\text{N, to the left} $$
    (b)
    Trial 2 ($8\,\text{N} = 8\,\text{N}$, resultant force $=0\,\text{N}$).
    (c)
    $$ 9 - 5 = 4\,\text{N, to the right} $$
    QUESTION 12 4 marks Criterion C
    Medium
    Force (N)2468
    Acceleration (m/s²)0.51.01.52.0

    This data was collected for a trolley of fixed mass, pulled with different forces.

    a. Describe the relationship between force and acceleration shown by this data.
    [2]
    b. Use the pattern in the data to predict the acceleration produced by a force of $10\,\text{N}$.
    [2]
    Show complete worked solution
    (a)
    As force increases, acceleration increases in direct proportion — doubling the force doubles the acceleration, and each extra $2\,\text{N}$ adds $0.5\,\text{m/s}^2$. Acceleration is directly proportional to the resultant force, consistent with Newton's Second Law.
    (b)
    Since acceleration is always $0.25$ times the force ($\dfrac{a}{F}=0.25$ in every row): $$ a = 10 \times 0.25 = 2.5\,\text{m/s}^2 $$
    QUESTION 13 5 marks Criterion C
    Medium
    Mass (kg)1248
    Acceleration (m/s²)8421

    This data was collected for a trolley pulled by the same, constant force each time.

    a. Show that these results are consistent with a constant applied force, by calculating the force for two of the rows.
    [3]
    b. Describe the relationship between mass and acceleration shown by this data, for a constant force.
    [2]
    Show complete worked solution
    (a)
    Using $F=ma$: for $m=1\,\text{kg}, a=8\,\text{m/s}^2$: $F = 1 \times 8 = 8\,\text{N}$. For $m=4\,\text{kg}, a=2\,\text{m/s}^2$: $F = 4 \times 2 = 8\,\text{N}$. Both rows give $F=8\,\text{N}$, confirming the applied force was the same throughout.
    (b)
    As mass increases, acceleration decreases. Specifically, doubling the mass halves the acceleration — mass and acceleration are inversely proportional when the force is kept constant.
    QUESTION 14 5 marks Criterion C
    Medium
    Force (N)12345
    Acceleration (m/s²)12745
    a. Identify the anomalous reading.
    [1]
    b. Explain your reasoning, using the pattern in the rest of the data.
    [2]
    c. State the mass of the trolley used, based on the consistent readings.
    [2]
    Show complete worked solution
    (a)
    The reading at $F=3\,\text{N}$ ($a=7\,\text{m/s}^2$) is anomalous.
    (b)
    In every other reading, acceleration equals the force exactly ($1,2,\_,4,5$). If this pattern had continued, the value at $F=3\,\text{N}$ should have been $3\,\text{m/s}^2$, not $7\,\text{m/s}^2$ — breaking the otherwise perfectly consistent pattern.
    (c)
    Since $a=F$ for every consistent reading, and $F=ma$, this means $m=1$ throughout. The trolley's mass is $1\,\text{kg}$.
    QUESTION 15 5 marks Criterion C
    Medium
    CarDriving force (N)Resistive force (N)
    A20002000
    B15001500
    C18001200
    a. Calculate the resultant force for each car.
    [3]
    b. State which car(s) are moving at constant velocity, and which is accelerating, referring to Newton's First Law.
    [2]
    Show complete worked solution
    (a)
    Car A: $2000-2000=0\,\text{N}$. Car B: $1500-1500=0\,\text{N}$. Car C: $1800-1200=600\,\text{N}$ (forward).
    (b)
    Cars A and B have a resultant force of zero (balanced forces), so by Newton's First Law they continue at whatever constant velocity they already have. Car C has a resultant force of $600\,\text{N}$ forward (unbalanced), so it is accelerating (speeding up) in the forward direction.
    QUESTION 16 8 marks Criterion C
    Hard
    Force (N)2468
    Acceleration of X (2 kg) (m/s²)1234
    Acceleration of Y (4 kg) (m/s²)0.511.52
    a. Calculate the acceleration produced by a $6\,\text{N}$ force on trolley X, and confirm it matches $F=ma$ using its mass.
    [2]
    b. A student claims "trolley Y always has exactly half the acceleration of trolley X for the same force." Evaluate this claim using the data in the table.
    [3]
    c. Explain why this pattern makes sense, given the two trolleys' masses.
    [3]
    Show complete worked solution
    (a)
    From the table, $a=3\,\text{m/s}^2$ at $F=6\,\text{N}$. Checking: $F=ma=2\,\text{kg}\times3\,\text{m/s}^2=6\,\text{N}$, which matches the given force — consistent.
    (b)
    Checking each force: at $2\,\text{N}$, X$=1$, Y$=0.5$ (half); at $4\,\text{N}$, X$=2$, Y$=1$ (half); at $6\,\text{N}$, X$=3$, Y$=1.5$ (half); at $8\,\text{N}$, X$=4$, Y$=2$ (half). In every case Y's acceleration is exactly half of X's, so the claim is fully supported by the data.
    (c)
    Trolley Y has exactly twice the mass of trolley X ($4\,\text{kg}$ vs $2\,\text{kg}$). Since $a=F/m$, for the same force $F$, doubling the mass exactly halves the acceleration — precisely the relationship the data shows, confirming Newton's Second Law's prediction that acceleration is inversely proportional to mass.
    QUESTION 17 7 marks Criterion C
    Hard

    A $3\,\text{kg}$ trolley is pulled by a constant resultant force of $6\,\text{N}$. Theoretically, $a=F/m=2\,\text{m/s}^2$. Five repeated trials measuring the actual acceleration gave: $1.85$, $2.10$, $1.95$, $2.30$, $1.90\,\text{m/s}^2$.

    a. Calculate the mean measured acceleration.
    [2]
    b. Calculate the percentage difference between the mean measured value and the theoretical value of $2\,\text{m/s}^2$.
    [2]
    c. Evaluate whether these results support Newton's Second Law, referring to both the mean value and the spread of the five results.
    [3]
    Show complete worked solution
    (a)
    $$ \text{mean} = \frac{1.85+2.10+1.95+2.30+1.90}{5} = \frac{10.10}{5} = 2.02\,\text{m/s}^2 $$
    (b)
    Difference $= 2.02 - 2.00 = 0.02\,\text{m/s}^2$. $$ \%\ \text{difference} = \frac{0.02}{2.00} \times 100 = 1\% $$
    (c)
    The mean measured acceleration ($2.02\,\text{m/s}^2$) is extremely close to the theoretical value of $2\,\text{m/s}^2$ (within $1\%$), strongly supporting Newton's Second Law. The five individual values do show some spread ($1.85$ to $2.30\,\text{m/s}^2$, a range of $0.45\,\text{m/s}^2$), likely due to experimental factors such as friction or small timing errors, but since they scatter fairly evenly above and below the theoretical value and the mean is so close to the prediction, the data still supports the law well.
    QUESTION 18 6 marks Criterion D
    Medium

    Newton's First Law explains why, in a car crash, an unrestrained passenger continues moving forward at the car's original speed even after the car itself has suddenly stopped — because no force is acting on the passenger to stop them at the same time as the car. Seatbelts are designed to provide this force.

    Discuss one benefit and one drawback of laws requiring seatbelt use.

    Show complete worked solution

    Benefit: A seatbelt applies a restraining (backward) force to the passenger's body at the same time the car decelerates, preventing the passenger from continuing to travel forward into the windscreen or dashboard due to their own inertia. This significantly reduces the severity of injuries and saves many lives in collisions, which is why seatbelt use is legally required in most countries.

    Drawback: Some passengers find seatbelts uncomfortable or restrictive on long journeys, and in rare cases (for example a vehicle submerged in water or on fire) a jammed seatbelt buckle can make it harder for a passenger to escape quickly. Specially designed harnesses and anchor points are also needed for very young children, adding cost and complexity for parents and vehicle manufacturers.

    QUESTION 19 6 marks Criterion D
    Medium

    Rocket engines work by Newton's Third Law: burning fuel is forced out of the rocket at high speed in one direction, and by Newton's Third Law an equal and opposite reaction force pushes the rocket in the other direction, allowing it to launch and travel through space, where there is no air or ground to push against.

    Discuss one benefit and one drawback of society's use of rockets built on this principle.

    Show complete worked solution

    Benefit: Rockets built on this principle have allowed the launch of satellites providing GPS navigation, weather forecasting, global communications, and internet access to remote areas, as well as enabling scientific research (such as studying climate change from orbit) that is not possible from the ground — technology that huge numbers of people now depend on daily.

    Drawback: Rocket launches are very expensive and use large amounts of fuel, releasing significant carbon dioxide and other emissions into the atmosphere. The growing number of launches has also left thousands of pieces of "space debris" (old rocket stages and satellite fragments) orbiting Earth at high speed, which pose a collision risk to working satellites and future spacecraft, including crewed missions.

    QUESTION 20 7 marks Criterion D
    Hard

    Government safety regulations require every new car model to pass crash tests, in which the forces experienced by crash-test dummies (calculated using Newton's Second Law, $F=ma$, from their measured deceleration) must stay below certain limits for the car to be sold legally.

    Evaluate the impact of these regulations on society, discussing both a benefit and a concern.

    Show complete worked solution

    Benefit: By legally requiring every new car to demonstrably limit the forces (and so the injury risk) experienced by occupants during a collision, these regulations have driven real, measurable improvements in car safety over recent decades, directly reducing deaths and serious injuries on the roads, and giving manufacturers a strong incentive to keep improving safety design rather than treating it as optional.

    Concern: Designing, building, and crash-testing cars to meet these standards adds significant cost to vehicle development, which is often passed on to buyers — this can make new, safer cars less affordable, particularly in lower-income countries or communities, meaning the safety benefit is not shared equally. Older, second-hand cars that predate stricter standards also remain on the road for many years, so the safety gap between the newest and oldest vehicles can be large.

    Friction, Air Resistance and Terminal Velocity 20 questions

    QUESTION 1 2 marks Criterion A
    Easy

    Define friction, stating what it always opposes.

    Show complete worked solution

    Friction is a contact force that acts between two surfaces (or between an object and a fluid such as air or water) in contact. It always opposes (acts against) the direction of relative motion (or attempted motion) between the surfaces.

    QUESTION 2 2 marks Criterion A
    Easy

    State two factors that affect the size of the friction force between two solid surfaces.

    Show complete worked solution

    (1) How rough or smooth the two surfaces are — rougher surfaces produce more friction. (2) How hard the surfaces are pressed together — a greater force pressing the surfaces together (e.g. a heavier object) produces more friction.

    QUESTION 3 2 marks Criterion A
    Easy
    weight air resistance

    The diagram shows a skydiver falling, with two forces acting on them: weight, and air resistance.

    a. Name the two forces shown acting on the skydiver.
    [1]
    b. State which force is bigger at this moment, and how you can tell from the diagram.
    [1]
    Show complete worked solution
    (a)
    Weight (gravity, acting downward) and air resistance (drag, acting upward).
    (b)
    Weight is bigger — its arrow is drawn longer than the air resistance arrow. This means there is a resultant force downward, so the skydiver is still accelerating (has not yet reached terminal velocity).
    QUESTION 4 4 marks Criterion A
    Medium

    A skydiver falls, speeding up as they fall.

    a. Explain what happens to air resistance as the skydiver's falling speed increases.
    [2]
    b. Explain why the skydiver eventually stops accelerating and falls at a constant ("terminal") velocity.
    [2]
    Show complete worked solution
    (a)
    As speed increases, air resistance increases too — more air is pushed out of the way, and pushed out of the way faster, every second.
    (b)
    As speed (and so air resistance) keeps increasing, air resistance eventually becomes exactly equal in size to the skydiver's weight. At this point, the upward and downward forces are balanced, giving a resultant force of zero — so by Newton's First Law, the skydiver stops accelerating and continues falling at a constant speed (terminal velocity).
    QUESTION 5 4 marks Criterion A
    Medium

    Friction and air resistance can both be reduced by design.

    a. Explain how oiling (lubricating) a bicycle chain reduces friction.
    [2]
    b. Explain how the streamlined (teardrop) shape of a racing car or a cyclist's helmet reduces air resistance.
    [2]
    Show complete worked solution
    (a)
    A thin layer of oil sits between the moving metal surfaces (the chain links and gears), reducing direct contact between the rough metal surfaces and letting them slide over each other more easily — this lowers the friction force between them.
    (b)
    A streamlined shape lets air flow smoothly around it with less turbulence, so less air is disturbed and less kinetic energy is transferred from the moving object to the air. This reduces the drag (air resistance) force compared to a flatter or more irregular shape moving at the same speed.
    QUESTION 6 4 marks Criterion A
    Medium

    A skydiver falls with their parachute closed, then opens it partway through the fall.

    a. Explain why opening the parachute causes the skydiver to decelerate suddenly.
    [2]
    b. Explain why the skydiver then reaches a new, much lower, terminal velocity with the parachute open.
    [2]
    Show complete worked solution
    (a)
    Opening the parachute suddenly increases the surface area facing the air, greatly increasing air resistance so that it becomes much larger than the skydiver's weight. This creates a large resultant force acting upward (opposing the fall), causing rapid deceleration.
    (b)
    As the skydiver decelerates, air resistance also decreases (since it depends on speed). This continues until air resistance again becomes exactly equal to weight — but because the parachute's much larger area means air resistance is already large even at low speeds, this new balance is reached at a much lower speed than before, giving a much lower terminal velocity.
    QUESTION 7 6 marks Criterion A
    Hard

    A steel ball bearing of mass $0.2\,\text{kg}$ is dropped from rest and falls through a tall column of oil. Assume $g=10\,\text{m/s}^2$.

    a. Calculate the ball's weight.
    [2]
    b. Explain what happens to the ball's acceleration during the fall, from the moment it is released until it reaches terminal velocity, referring to the changing size of the resistive (drag) force from the oil.
    [2]
    c. State the resultant force on the ball once it reaches terminal velocity, and explain your reasoning.
    [2]
    Show complete worked solution
    (a)
    $$ W = mg = 0.2 \times 10 $$

    Answer: $W = 2\,\text{N}$

    (b)
    At the start (low speed), the drag force is small compared to the ball's weight ($2\,\text{N}$), so there is a large resultant force downward and the ball accelerates quickly. As speed increases, drag increases too, so the resultant force gets smaller and smaller — the ball's acceleration decreases (it still speeds up, but by less and less each second) until drag equals $2\,\text{N}$.
    (c)
    The resultant force is $0\,\text{N}$. At terminal velocity, the drag force has increased until it exactly equals the ball's weight ($2\,\text{N}$), so the upward and downward forces are balanced, giving zero resultant force and constant velocity (zero acceleration).
    QUESTION 8 6 marks Criterion B
    Medium

    A student wants to compare the friction between a wooden block and different surface materials.

    a. State the independent variable, the dependent variable, and one variable that must be controlled.
    [3]
    b. Describe a method using a newton-meter, a wooden block, and different surface materials to compare friction.
    [3]
    Show complete worked solution
    (a)
    Independent variable: type of surface material (e.g. carpet, bare wood, sandpaper). Dependent variable: force needed to just start the block moving (or to pull it at a constant speed), measured with a newton-meter, as a measure of friction. Controlled variable: the block used (same weight) and the distance it is dragged, kept the same across all trials.
    (b)
    Attach a newton-meter to the wooden block and place it on the first surface (e.g. bare table). Slowly pull the newton-meter horizontally and record the reading at the moment the block just starts to move. Repeat $3$ times and calculate a mean. Then place the same block on each different surface material in turn (e.g. carpet, sandpaper) and repeat the same procedure, comparing the mean force needed for each surface — a larger force needed indicates a rougher surface with more friction.
    QUESTION 9 6 marks Criterion B
    Medium

    A student wants to investigate how the surface area of a piece of paper affects the time it takes to fall a fixed height, using the same paper cut into squares of different sizes (so the material and thickness stay the same).

    a. State the independent and dependent variables, and one variable that must be controlled.
    [3]
    b. Describe how the student could ensure a fair test when dropping each paper square, and how they could check that the paper had reached terminal velocity.
    [3]
    Show complete worked solution
    (a)
    Independent variable: surface area of the paper square (e.g. $5\,\text{cm}\times5\,\text{cm}$, $10\,\text{cm}\times10\,\text{cm}$, $15\,\text{cm}\times15\,\text{cm}$). Dependent variable: time taken to fall a fixed height. Controlled variable: the drop height and the material/thickness of the paper — kept the same for every square.
    (b)
    Drop each paper square from the same fixed height (e.g. $2\,\text{m}$), released from rest with the flat side facing down each time, using a consistent release method (e.g. held flat between two fingers and let go, not thrown). Time the fall with a stopwatch, repeating each size $3$ times and taking a mean. To check terminal velocity was reached, use two markers along the drop (e.g. at $1\,\text{m}$ and $2\,\text{m}$) and confirm the time taken to fall the last section equals the time for an equal earlier section — showing the paper was falling at a constant, no-longer-increasing speed by that point.
    QUESTION 10 8 marks Criterion B
    Hard

    A student times a marble falling through a tall tube of golden syrup to investigate terminal velocity, marking two points on the tube $20\,\text{cm}$ apart near the bottom (assumed to be within the terminal-velocity region) and timing the marble between them by eye with a stopwatch.

    a. Explain one reason why timing "by eye" in this way might not accurately capture the exact moment the marble passes each mark, and how this could affect the calculated terminal velocity.
    [3]
    b. The student assumes the marble has definitely reached terminal velocity by the time it reaches the first mark, without checking. Explain how the student could check this assumption using data from an earlier section of the tube.
    [3]
    c. Suggest one modification to the experimental setup that would make the timing more precise.
    [2]
    Show complete worked solution
    (a)
    Judging exactly when a small, possibly fast-moving marble passes a mark by eye is subject to human reaction time and misjudging its exact position, especially in a viscous liquid where the marble may not be perfectly visible. This random error would make the measured time between marks sometimes too long and sometimes too short between repeats, making the calculated speed (and so terminal velocity) less precise and reliable.
    (b)
    The student could mark a second, earlier interval further up the tube (e.g. the $20\,\text{cm}$ section just above the first) and time the marble over that section too. If the calculated speed for both sections is the same (within experimental error), this shows the marble's speed was no longer increasing over that stretch — i.e. it had genuinely reached a constant (terminal) velocity by the first mark, rather than still accelerating.
    (c)
    Use light gates positioned at the marks, connected to an electronic timer, so the exact moment the marble (or a small flag attached to it) passes each point is detected automatically and precisely, removing human reaction-time and judgement error from the timing.
    QUESTION 11 3 marks Criterion C
    Easy
    SurfaceIceWoodCarpetSandpaper
    Force needed (N)0.52.03.55.0
    a. Which surface produces the most friction?
    [1]
    b. State the general pattern shown by the data linking surface roughness and friction force.
    [1]
    c. Suggest why ice needs such a small force compared to the others.
    [1]
    Show complete worked solution
    (a)
    Sandpaper — it needs the largest force ($5.0\,\text{N}$) to slide the block.
    (b)
    The rougher the surface, the greater the force needed to slide the block — i.e. rougher surfaces produce more friction.
    (c)
    Ice is extremely smooth, so there is very little for the two surfaces to grip or catch on — meaning very little friction opposes the block's motion.
    QUESTION 12 4 marks Criterion C
    Medium
    Time (s)0123456
    Speed (m/s)091621242525

    This table records the speed of a falling object.

    a. At what time does the object appear to reach terminal velocity? Explain how the data shows this.
    [2]
    b. Calculate the object's acceleration between $t=0$ and $t=1\,\text{s}$, and between $t=4$ and $t=5\,\text{s}$, then use these to describe how acceleration changes as terminal velocity is approached.
    [2]
    Show complete worked solution
    (a)
    At $t=5\,\text{s}$ (continuing at $t=6\,\text{s}$), since speed stays constant at $25\,\text{m/s}$ for two consecutive readings, showing the acceleration has become zero.
    (b)
    $0$–$1\,\text{s}$: $a=\dfrac{9-0}{1}=9\,\text{m/s}^2$. $4$–$5\,\text{s}$: $a=\dfrac{25-24}{1}=1\,\text{m/s}^2$. Acceleration decreases sharply as terminal velocity is approached — from $9\,\text{m/s}^2$ near the start to just $1\,\text{m/s}^2$ close to reaching terminal velocity, before becoming $0\,\text{m/s}^2$ once it is reached.
    QUESTION 13 5 marks Criterion C
    Medium
    0 1 2 3 4 5 6 7 0 10 20 30 40 Time (s) Speed (m/s)

    The graph shows the speed of a skydiver during the first $7\,\text{s}$ of a fall.

    a. Describe the shape of the graph, and what it shows about the skydiver's acceleration over time.
    [2]
    b. Estimate the skydiver's terminal velocity from the graph, and state how you can tell from the graph's shape.
    [1]
    c. Explain, in terms of forces, why the graph becomes flat rather than continuing to curve upward forever.
    [2]
    Show complete worked solution
    (a)
    The graph rises steeply at first, then curves and flattens out, becoming horizontal from about $t=5\,\text{s}$ onward. This shows the skydiver's acceleration is greatest at the start (steepest gradient) and gradually decreases as speed increases, until acceleration becomes zero (flat line) once terminal velocity is reached.
    (b)
    Approximately $36\,\text{m/s}$; the line becomes horizontal (flat, constant height) from about $t=5\,\text{s}$ onward, showing speed is no longer changing.
    (c)
    The graph flattens because, by that point, air resistance has increased (as speed increased) until it exactly balances the skydiver's weight. With the forces balanced (zero resultant force), there is no more acceleration, so speed stops increasing and the graph becomes a horizontal line.
    QUESTION 14 5 marks Criterion C
    Medium
    ShapeCoffee filter (flat)Coffee filter (crumpled into a ball)
    Time to fall 2 m (s)2.40.6

    Both filters have the same mass.

    a. Calculate the average speed of each filter during the fall.
    [2]
    b. Explain, in terms of air resistance, why the crumpled filter falls so much faster despite having the same mass as the flat one.
    [3]
    Show complete worked solution
    (a)
    Flat: $v=\dfrac{2}{2.4}=0.833\,\text{m/s}$ (3 s.f.). Crumpled: $v=\dfrac{2}{0.6}=3.33\,\text{m/s}$ (3 s.f.).
    (b)
    Both filters have the same weight (same mass), but the flat filter has a much larger surface area facing the air as it falls, so it experiences a much larger air resistance force at any given speed — reaching a much lower terminal velocity. The crumpled filter has a much smaller surface area, so it experiences far less air resistance at a given speed, and must reach a much higher speed before air resistance grows large enough to balance its (equal) weight — so it reaches a higher terminal velocity and falls faster.
    QUESTION 15 5 marks Criterion C
    Medium
    Time (s)012345
    Speed (m/s)0814181521
    a. Identify the anomalous reading.
    [1]
    b. Explain why this reading must be an error, using an idea about the physics of the fall (not just the numeric pattern).
    [2]
    c. Suggest what the value at $t=4\,\text{s}$ should probably have been, based on the pattern in the surrounding readings.
    [2]
    Show complete worked solution
    (a)
    The reading at $t=4\,\text{s}$ ($15\,\text{m/s}$) is anomalous.
    (b)
    Before terminal velocity is reached, the falling object's resultant force is always downward (weight still greater than air resistance), so it must keep speeding up — its speed should keep increasing every second, never decrease. A drop from $18\,\text{m/s}$ to $15\,\text{m/s}$ is not physically possible in this situation, so it must be a measurement or recording error.
    (c)
    The differences before the anomaly are decreasing steadily ($8, 6, 4\,\text{m/s}$ per second); continuing this pattern (next difference $\approx2$), the value at $t=4\,\text{s}$ should probably have been close to $20\,\text{m/s}$ ($18+2$), consistent with speed still rising but by a smaller amount each second as terminal velocity is approached.
    QUESTION 16 8 marks Criterion C
    Hard
    Parachute diameter (cm)20406080
    Terminal velocity (m/s)8.56.04.94.2
    a. Describe the relationship between parachute diameter and terminal velocity shown by the data.
    [2]
    b. A student concludes "doubling the parachute's diameter always halves the terminal velocity." Evaluate this claim using the data.
    [3]
    c. Explain, in terms of forces, why a larger parachute produces a lower terminal velocity.
    [3]
    Show complete worked solution
    (a)
    As parachute diameter increases, terminal velocity decreases — a larger parachute produces a lower terminal (falling) speed. However, the decrease is not proportional: a much bigger increase in diameter is needed to produce the same-sized drop in terminal velocity as diameter gets larger (e.g. doubling diameter from $20$ to $40\,\text{cm}$ drops speed by $2.5\,\text{m/s}$, but doubling again from $40$ to $80\,\text{cm}$ drops it by only a further $1.8\,\text{m/s}$).
    (b)
    Checking: $20\,\text{cm}\to8.5\,\text{m/s}$; doubling to $40\,\text{cm}$ gives $6.0\,\text{m/s}$, not $4.25\,\text{m/s}$ (half of $8.5$) — so doubling diameter clearly does not halve terminal velocity here. Similarly, $40\,\text{cm}\to6.0\,\text{m/s}$ and doubling to $80\,\text{cm}$ gives $4.2\,\text{m/s}$, again not half (which would be $3.0\,\text{m/s}$). The claim is not supported by the data.
    (c)
    A larger parachute has a greater surface area, so it experiences a greater air resistance force for a given falling speed than a smaller parachute would. This means air resistance grows to equal the skydiver's weight (the balance needed for terminal velocity) at a much lower speed for a larger parachute than for a smaller one — so a bigger parachute results in a lower terminal velocity.
    QUESTION 17 7 marks Criterion C
    Hard
    Trial12345
    Time to fall 1.5 m (s)1.421.381.511.401.44

    Five trials measured the time for a paper cone to fall a fixed $1.5\,\text{m}$ distance, assumed to be at terminal velocity.

    a. Calculate the mean time.
    [2]
    b. Calculate the terminal velocity using the mean time.
    [2]
    c. One reading ($1.51\,\text{s}$) is noticeably higher than the rest. Suggest a possible experimental reason for this single higher reading (not necessarily a mistake to discard), and explain how repeating trials and averaging helps deal with this kind of natural variation.
    [3]
    Show complete worked solution
    (a)
    $$ \text{mean} = \frac{1.42+1.38+1.51+1.40+1.44}{5} = \frac{7.15}{5} = 1.43\,\text{s} $$
    (b)
    $$ v = \frac{1.5}{1.43} = 1.05\,\text{m/s} \ (\text{3 s.f.}) $$
    (c)
    A possible reason is that on that trial the cone was released with a very slight sideways tilt or spin, or caught a small air current, causing it to take a very slightly longer, less direct path down — a realistic, natural source of small random variation rather than necessarily a mistake. Repeating the trial multiple times and averaging reduces the effect of any one such unusual trial on the final result, since random variations tend to happen in both directions (too fast and too slow) and largely cancel out when averaged, giving a mean value more representative of the true terminal velocity than any single trial alone.
    QUESTION 18 6 marks Criterion D
    Medium

    Car and truck manufacturers increasingly design vehicles with more streamlined (aerodynamic) shapes to reduce air resistance, which lowers fuel consumption and CO2 emissions, especially at higher motorway speeds where air resistance has the greatest effect.

    Discuss one benefit and one drawback of prioritising streamlined design in vehicles.

    Show complete worked solution

    Benefit: Reducing air resistance directly reduces the driving force (and so fuel) needed to maintain speed, especially on motorways — lowering fuel costs for drivers and reducing the CO2 and other emissions released per kilometre travelled, helping reduce vehicles' overall contribution to climate change and air pollution.

    Drawback: Prioritising a streamlined shape can conflict with other practical needs — for example, delivery vans and trucks need large, boxy shapes to maximise the space available for cargo, and a more streamlined design may reduce this usable space. Achieving very streamlined shapes can also require expensive materials and specialist engineering/wind-tunnel testing, increasing the manufacturing cost of the vehicle.

    QUESTION 19 6 marks Criterion D
    Medium

    Car tyres are designed with a tread pattern (grooves) that increases friction/grip between the tyre and a wet road, by channelling water away from the contact area.

    Discuss one benefit and one drawback of tyre tread design choices.

    Show complete worked solution

    Benefit: A good tread pattern significantly increases grip in wet conditions by clearing water from between the tyre and the road, reducing the risk of "aquaplaning" (where a layer of water lifts the tyre off the road, drastically reducing friction and control). This directly improves road safety, particularly for braking distances in rain.

    Drawback: Tyres with a more aggressive tread designed for maximum grip typically also increase rolling friction/resistance with the road, meaning the engine has to work harder (using more fuel) to keep the car moving at a constant speed compared to smoother, low-resistance tyres. Tyre tread also wears down over time through friction, releasing small rubber/microplastic particles into the environment and requiring tyres to be replaced periodically, adding cost and waste.

    QUESTION 20 7 marks Criterion D
    Hard

    Advances in lightweight, high-strength fabrics have allowed engineers to design smaller, lighter parachutes that still slow a falling object to a safe terminal velocity — used for everything from recreational skydiving to landing spacecraft safely back on Earth.

    Evaluate the impact of this technology, discussing both a benefit and a concern.

    Show complete worked solution

    Benefit: Lighter, more compact parachute systems take up less mass and space, which is especially valuable in spacecraft (where every kilogram of payload is extremely expensive to launch), and stronger, more reliable fabrics reduce the risk of parachute failure — giving a safer, more controlled and predictable terminal velocity for landing astronauts, cargo, or Mars rovers safely.

    Concern: This advanced fabric technology, and the engineering and testing needed to certify a parachute system as safe, is very expensive, meaning access to the most reliable, cutting-edge parachute systems tends to be limited to well-funded space agencies and companies rather than being available to all. A parachute is also a single point of failure in situations like spacecraft re-entry — if a small manufacturing fault or unexpected tear affects the fabric's strength, the consequences of a malfunction at the extreme speeds and heights involved (failing to reach a safe terminal velocity in time) can be catastrophic, so extremely rigorous (and costly) testing is essential.