Which physical quantity is equal to the gradient of a displacement-time graph?
Show complete worked solution
Correct answer: B
Write the definition of velocity.
$$v=\frac{\Delta s}{\Delta t}$$
A graph gradient is change in the vertical quantity divided by change in the horizontal quantity. For a displacement-time graph this is $\Delta s/\Delta t$.
Therefore the gradient represents velocity.
$$\boxed{\text{B}}$$
QUESTION 2[1 mark]
Multiple choiceFoundationCalculatorAbout 1 min
A car's velocity increases uniformly from $5.0\,\mathrm{m\,s^{-1}}$ to $17.0\,\mathrm{m\,s^{-1}}$ in $4.0\,\mathrm{s}$. What is its acceleration?
Show complete worked solution
Correct answer: B
Use the definition of acceleration.
$$a=\frac{v-u}{t}$$
Substitute the values.
$$a=\frac{17.0-5.0}{4.0}=3.0\,\mathrm{m\,s^{-2}}$$
Match this with the options.
$$\boxed{\text{B}}$$
QUESTION 3[1 mark]
Multiple choiceFoundationCalculatorAbout 1 min
An object moves with constant positive velocity. Which description matches its displacement-time graph?
Show complete worked solution
Correct answer: A
The gradient of a displacement-time graph is velocity.
Constant velocity requires a constant gradient, so the graph must be a straight line.
Positive velocity requires a positive gradient.
$$\boxed{\text{A}}$$
QUESTION 4[1 mark]
Multiple choiceFoundationCalculatorAbout 1 min
A stone is released from rest and falls $19.6\,\mathrm{m}$. Air resistance is negligible and $g=9.8\,\mathrm{m\,s^{-2}}$. How long does it fall?
Show complete worked solution
Correct answer: C
Since the stone is released from rest, $u=0$.
Use $s=ut+\tfrac12at^2$.
$$19.6=0+\frac12(9.8)t^2$$
$$t^2=4.00$$
Time is positive.
$$t=2.00\,\mathrm{s}$$
$$\boxed{\text{C}}$$
QUESTION 5[1 mark]
Multiple choiceStandardCalculatorAbout 1.5 min
A projectile moves through the air with negligible air resistance. At its highest point, which statement is correct?
Show complete worked solution
Correct answer: D
Resolve velocity into horizontal and vertical components.
At the highest point the vertical component is momentarily zero, but the horizontal component remains non-zero.
Gravity still acts vertically downward, so acceleration is $g$ downward.
$$\boxed{\text{D}}$$
QUESTION 6[1 mark]
Multiple choiceStandardCalculatorAbout 2 min
A speed is calculated using $v=d/t$, where $d=(12.0\pm0.2)\,\mathrm{m}$ and $t=(3.00\pm0.06)\,\mathrm{s}$. Which result correctly includes the maximum absolute uncertainty?
Show complete worked solution
Correct answer: B
Calculate the speed.
$$v=\frac{12.0}{3.00}=4.00\,\mathrm{m\,s^{-1}}$$
For division, add percentage uncertainties.
$$\frac{\Delta v}{v}=\frac{0.2}{12.0}+\frac{0.06}{3.00}=0.0367$$
Convert to absolute uncertainty.
$$\Delta v=4.00(0.0367)=0.147\,\mathrm{m\,s^{-1}}\approx0.15\,\mathrm{m\,s^{-1}}$$
$$\boxed{v=(4.00\pm0.15)\,\mathrm{m\,s^{-1}}}$$
QUESTION 7[4 marks]
Short responseStandardCalculatorAbout 4 min
A car is travelling at $24\,\mathrm{m\,s^{-1}}$ when its brakes produce a constant acceleration of $-6.0\,\mathrm{m\,s^{-2}}$. Determine the time and distance required for the car to stop.
Show complete worked solution
Find the stopping time.
$$v=u+at$$
$$0=24-6.0t$$
$$t=4.0\,\mathrm{s}$$
Find the braking distance.
$$v^2=u^2+2as$$
$$0=24^2+2(-6.0)s$$
$$s=\frac{576}{12}=48\,\mathrm{m}$$
$$\boxed{t=4.0\,\mathrm{s},\quad s=48\,\mathrm{m}}$$
Final answer: $t=4.0\,\mathrm{s}$ and $s=48\,\mathrm{m}$
QUESTION 8[5 marks]
Short responseStandardCalculatorAbout 5 min
A vehicle starts from rest and accelerates uniformly to $12\,\mathrm{m\,s^{-1}}$ in $4.0\,\mathrm{s}$. It travels at this speed for $6.0\,\mathrm{s}$ and then decelerates uniformly to rest in $3.0\,\mathrm{s}$. Determine the total distance travelled.
Show complete worked solution
The distance is the area under the velocity-time graph.
Add the areas.
$$s=24+72+18=\boxed{114\,\mathrm{m}}$$
Final answer: $114\,\mathrm{m}$
QUESTION 9[5 marks]
Short responseAppliedCalculatorAbout 5 min
A driver travels at $20\,\mathrm{m\,s^{-1}}$. The reaction time is $0.70\,\mathrm{s}$. After the brakes are applied, the car decelerates uniformly at $5.0\,\mathrm{m\,s^{-2}}$. Calculate the total stopping distance.
Show complete worked solution
Reaction distance. During the reaction time the speed is unchanged.
$$s_r=vt=(20)(0.70)=14\,\mathrm{m}$$
Total stopping distance.
$$s_{total}=14+40=\boxed{54\,\mathrm{m}}$$
Final answer: $54\,\mathrm{m}$
QUESTION 10[6 marks]
Short responseStandardCalculatorAbout 6 min
A ball is projected vertically upward from ground level at $18.0\,\mathrm{m\,s^{-1}}$. Air resistance is negligible. Use $g=9.81\,\mathrm{m\,s^{-2}}$ to determine (i) the time to maximum height, (ii) the maximum height, and (iii) the total flight time.
Show complete worked solution
Take upward as positive, so $a=-9.81\,\mathrm{m\,s^{-2}}$.
Time to maximum height. At the top, $v=0$.
$$0=18.0-9.81t$$
$$t_{up}=1.83\,\mathrm{s}$$
Maximum height.
$$v^2=u^2+2as$$
$$0=18.0^2-2(9.81)h$$
$$h=16.5\,\mathrm{m}$$
Total flight time. The motion is symmetric because launch and landing are at the same level.
$$t_{total}=2t_{up}=3.67\,\mathrm{s}$$
$$\boxed{t_{up}=1.83\,\mathrm{s},\ h=16.5\,\mathrm{m},\ t_{total}=3.67\,\mathrm{s}}$$
Final answer: $1.83\,\mathrm{s}$; $16.5\,\mathrm{m}$; $3.67\,\mathrm{s}$
QUESTION 11[4 marks]
Short responseStandardCalculatorAbout 4 min
A train moves east at $22.0\,\mathrm{m\,s^{-1}}$. A passenger walks at $1.50\,\mathrm{m\,s^{-1}}$ relative to the train. Determine the passenger's velocity relative to the ground when walking (a) east and (b) west.
Show complete worked solution
Take east as positive. Use $v_{PG}=v_{PT}+v_{TG}$.
(a) Walking east:
$$v_{PG}=1.50+22.0=\boxed{23.5\,\mathrm{m\,s^{-1}}\text{ east}}$$
(b) Walking west: $v_{PT}=-1.50\,\mathrm{m\,s^{-1}}$.
$$v_{PG}=-1.50+22.0=\boxed{20.5\,\mathrm{m\,s^{-1}}\text{ east}}$$
Final answer: $23.5\,\mathrm{m\,s^{-1}}$ east; $20.5\,\mathrm{m\,s^{-1}}$ east
QUESTION 12[6 marks]
Short responseAppliedCalculatorAbout 6 min
A trolley travels $d=(50.0\pm0.2)\,\mathrm{m}$ in $t=(4.00\pm0.05)\,\mathrm{s}$. Determine its average speed and maximum absolute uncertainty.
Show complete worked solution
Calculate the average speed.
$$v=\frac{d}{t}=\frac{50.0}{4.00}=12.5\,\mathrm{m\,s^{-1}}$$
Add fractional uncertainties for division.
$$\frac{\Delta v}{v}=\frac{0.2}{50.0}+\frac{0.05}{4.00}=0.0040+0.0125=0.0165$$
This is $1.65\%$.
Convert to absolute uncertainty.
$$\Delta v=(12.5)(0.0165)=0.206\,\mathrm{m\,s^{-1}}\approx0.2\,\mathrm{m\,s^{-1}}$$
The table shows the velocity of a cart at one-second intervals. Determine the acceleration, the displacement from $0$ to $5.0\,\mathrm{s}$, and an equation for $v(t)$.
Show complete worked solution
Find the acceleration from the gradient.
$$a=\frac{17-2}{5-0}=3.0\,\mathrm{m\,s^{-2}}$$
Find displacement from the area under the velocity-time graph.
$$s=\frac{u+v}{2}t=\frac{2+17}{2}(5)=47.5\,\mathrm{m}$$
Write the velocity equation.
$$v=u+at=2.0+3.0t$$
$$\boxed{a=3.0\,\mathrm{m\,s^{-2}},\quad s=47.5\,\mathrm{m},\quad v=2.0+3.0t}$$
Final answer: $a=3.0\,\mathrm{m\,s^{-2}}$; $s=47.5\,\mathrm{m}$; $v=2.0+3.0t$
QUESTION 14[7 marks]
Data-basedAppliedCalculatorAbout 8 min
$t/\mathrm{s}$
0
1
2
3
4
$x/\mathrm{m}$
0
2
8
18
32
The position of an object is recorded as shown. Use the data to show that the acceleration is constant, determine its value, and predict the instantaneous velocity at $t=3.0\,\mathrm{s}$.
Show complete worked solution
Calculate average velocities over each one-second interval.
$$v_{0-1}=2,\quad v_{1-2}=6,\quad v_{2-3}=10,\quad v_{3-4}=14\,\mathrm{m\,s^{-1}}$$
These values apply at midpoint times $0.5,1.5,2.5,3.5\,\mathrm{s}$.
Find the change in velocity per second.
$$a=\frac{6-2}{1.5-0.5}=4.0\,\mathrm{m\,s^{-2}}$$
The same increase occurs in every interval, so acceleration is constant.
Identify the motion equation. The data fit $x=2t^2$. Comparing with $x=ut+\tfrac12at^2$ gives $u=0$ and $a=4.0\,\mathrm{m\,s^{-2}}$.
Final answer: $a=4.0\,\mathrm{m\,s^{-2}}$ and $v(3.0\,\mathrm{s})=12\,\mathrm{m\,s^{-1}}$
QUESTION 15[8 marks]
Extended responseAdvancedCalculatorAbout 9 min
An electric train starts from rest and accelerates uniformly at $2.50\,\mathrm{m\,s^{-2}}$ until it reaches $30.0\,\mathrm{m\,s^{-1}}$. It maintains this speed for $18.0\,\mathrm{s}$ and then brakes uniformly at $5.00\,\mathrm{m\,s^{-2}}$ to rest. Determine the total distance, total time and average speed for the journey.
Average speed.
$$\bar v=\frac{810}{36.0}=\boxed{22.5\,\mathrm{m\,s^{-1}}}$$
Final answer: $810\,\mathrm{m}$; $36.0\,\mathrm{s}$; $22.5\,\mathrm{m\,s^{-1}}$
QUESTION 16[6 marks]
Extended responseStandardCalculatorAbout 7 min
Runner A passes a marker at a constant speed of $15\,\mathrm{m\,s^{-1}}$. Runner B passes the same marker $4.0\,\mathrm{s}$ later at a constant speed of $25\,\mathrm{m\,s^{-1}}$. Determine when and where B catches A, measured from the instant A passes the marker.
Show complete worked solution
Let $t$ be the time after A passes the marker.
Write A's displacement.
$$x_A=15t$$
Write B's displacement. B moves for $t-4.0$ seconds.
$$x_B=25(t-4.0)$$
Equate displacements at the catch.
$$15t=25(t-4.0)$$
$$15t=25t-100$$
$$t=10.0\,\mathrm{s}$$
Find the position.
$$x=15(10.0)=\boxed{150\,\mathrm{m}}$$
B catches A $10.0\,\mathrm{s}$ after A passes the marker, or $6.0\,\mathrm{s}$ after B passes it.
Final answer: $10.0\,\mathrm{s}$ after A; $150\,\mathrm{m}$ from the marker
QUESTION 17[8 marks]
Multi-partAdvancedCalculatorAbout 10 min
A lift starts from rest and accelerates upward at $1.20\,\mathrm{m\,s^{-2}}$ for $2.50\,\mathrm{s}$. It then moves at constant speed for $5.00\,\mathrm{s}$ before decelerating uniformly to rest in $2.00\,\mathrm{s}$.
a. Calculate the maximum speed.[2]
b. Calculate the total vertical distance travelled.[4]
c. Determine the average speed for the complete motion.[2]
A ball is launched horizontally at $12.0\,\mathrm{m\,s^{-1}}$ from a cliff $45.0\,\mathrm{m}$ above level ground. Air resistance is negligible. Use $g=9.81\,\mathrm{m\,s^{-2}}$.
a. Calculate the time taken to reach the ground.[2]
b. Determine the horizontal distance travelled.[2]
c. Determine the magnitude and direction of the velocity immediately before impact.[3]
Show complete worked solution
(a)
Vertical motion starts with $u_y=0$.
$$45.0=\tfrac12(9.81)t^2$$
$$t=\sqrt{\frac{90.0}{9.81}}=\boxed{3.03\,\mathrm{s}}$$
Final answer: $3.03\,\mathrm{s}$
(b)
Horizontal velocity is constant.
$$x=v_xt=(12.0)(3.03)=\boxed{36.3\,\mathrm{m}}$$
Final answer: $36.3\,\mathrm{m}$
(c)
$$v_y=gt=(9.81)(3.03)=29.7\,\mathrm{m\,s^{-1}}$$
$$v=\sqrt{12.0^2+29.7^2}=32.0\,\mathrm{m\,s^{-1}}$$
$$\theta=\tan^{-1}\left(\frac{29.7}{12.0}\right)=\boxed{68.0^\circ\text{ below the horizontal}}$$
Final answer: $32.0\,\mathrm{m\,s^{-1}}$ at $68.0^\circ$ below horizontal
QUESTION 19[8 marks]
Multi-partStandardCalculatorAbout 10 min
A boat must cross a river $120\,\mathrm{m}$ wide. Relative to the water, it travels due north at $5.0\,\mathrm{m\,s^{-1}}$. The river current is $3.0\,\mathrm{m\,s^{-1}}$ due east.
a. State the east and north components of the boat's velocity relative to the ground.[1]
b. Calculate the crossing time.[2]
c. Calculate the downstream displacement.[2]
d. Determine the magnitude and direction of the boat's velocity relative to the ground.[3]
$$v=\sqrt{5.0^2+3.0^2}=5.83\,\mathrm{m\,s^{-1}}$$
$$\theta=\tan^{-1}\left(\frac{3.0}{5.0}\right)=\boxed{31.0^\circ\text{ east of north}}$$
Final answer: $5.83\,\mathrm{m\,s^{-1}}$ at $31.0^\circ$ east of north
QUESTION 20[12 marks]
Multi-partAdvancedCalculatorAbout 14 min
A car travels at $28.0\,\mathrm{m\,s^{-1}}$ on a wet road. The driver's reaction time is $0.65\,\mathrm{s}$. After braking begins, the car decelerates at $4.0\,\mathrm{m\,s^{-2}}$ until its speed is $14.0\,\mathrm{m\,s^{-1}}$, then at $7.0\,\mathrm{m\,s^{-2}}$ until it stops.
a. Calculate the reaction distance.[2]
b. Determine the time and distance for the first braking stage.[4]
c. Determine the time and distance for the second braking stage.[3]
d. Calculate the total stopping distance and total stopping time, including the reaction phase.[3]
Final answer: $105.7\,\mathrm{m}$ and $6.15\,\mathrm{s}$
QUESTION 21[1 mark]
Multiple choiceStandardCalculatorAbout 2 min
A drone follows a circular path of radius $r$ from A to B as shown. The angle subtended at the centre is $120^\circ$. What is the magnitude of the drone's displacement?
Show complete worked solution
Correct answer: D
Displacement is the straight-line chord AB, not the distance travelled along the arc.
For central angle $120^\circ$, $$d=2r\sin\left(\frac{120^\circ}{2}\right)$$
$$d=2r\sin60^\circ=2r\left(\frac{\sqrt3}{2}\right)=\sqrt3r$$ Therefore option D is correct.
QUESTION 22[1 mark]
Multiple choiceAppliedCalculatorAbout 2 min
A shuttle accelerates uniformly between timing gates P and Q. It passes P at time $t_1$ with speed $u$, passes Q at $t_2$ with speed $v$, and the separation is $s$. Which expressions are valid? I. $a=(v-u)/(t_2-t_1)$ II. $a=(v^2-u^2)/(2s)$ III. $s=\tfrac12(u+v)(t_2-t_1)$
Show complete worked solution
Correct answer: D
Constant acceleration gives $$v=u+a\Delta t\Rightarrow a=\frac{v-u}{\Delta t}$$ so I is valid.
Using $v^2=u^2+2as$ gives $$a=\frac{v^2-u^2}{2s}$$ so II is valid.
Average speed is $(u+v)/2$, hence $$s=\frac{u+v}{2}\Delta t$$ so III is valid. Therefore option D is correct.
QUESTION 23[1 mark]
Multiple choiceAppliedCalculatorAbout 2 min
A supply package is released from rest at height $H$. At the same instant, a marker is launched vertically upward from the ground with speed $u$. Air resistance is negligible. At what time are they at the same height?
Show complete worked solution
Correct answer: C
Package height is $$y_1=H-\frac12gt^2$$
Marker height is $$y_2=ut-\frac12gt^2$$
Set $y_1=y_2$. The identical gravitational terms cancel: $$H=ut$$ Therefore $$t=\frac{H}{u}$$ and option C is correct.
QUESTION 24[1 mark]
Multiple choiceAppliedCalculatorAbout 2 min
A ball moves upward and to the right through still air at point P. Air resistance is significant. Which description of its acceleration components at P is correct?
Show complete worked solution
Correct answer: B
Drag acts opposite the velocity, so it has a leftward component and a downward component.
The leftward drag produces a non-zero horizontal acceleration.
Vertically, both weight and the downward drag act downward, so the vertical acceleration magnitude is greater than $g$. Therefore option B is correct.
QUESTION 25[1 mark]
Multiple choiceStandardCalculatorAbout 2 min
The velocity-time graph is a downward-opening parabola that starts and ends at zero velocity over $6.0\,\mathrm{s}$. Which description best matches the acceleration-time graph?
Show complete worked solution
Correct answer: A
Acceleration is the gradient of a velocity-time graph.
The parabola initially has a positive gradient, has zero gradient at its maximum, and later has a negative gradient.
The derivative of a quadratic curve is a straight line, so acceleration decreases linearly through zero. Option A is correct.
QUESTION 26[1 mark]
Multiple choiceStandardCalculatorAbout 2 min
An object has the positive acceleration-time graph shown. It starts with velocity $-4.0\,\mathrm{m\,s^{-1}}$. Which statement about its velocity from $0$ to $T$ must be correct?
Show complete worked solution
Correct answer: B
The acceleration is positive for $0
Acceleration is largest near the middle, so velocity increases most rapidly there.
Because acceleration varies, velocity is not linear. Therefore option B is correct.
QUESTION 27[1 mark]
Multiple choiceStandardCalculatorAbout 2 min
A probe is launched from a platform above level ground. To determine its horizontal landing distance without air resistance, which quantities must be known? I. Initial horizontal velocity component II. Initial vertical velocity component III. Platform height
Show complete worked solution
Correct answer: D
Horizontal distance is $x=u_xt$, so the horizontal component I is required.
Flight time follows from vertical motion: $$0=H+u_yt-\frac12gt^2$$ This requires both vertical component II and height III.
All three quantities are required. Option D is correct.
QUESTION 28[1 mark]
Multiple choiceAppliedCalculatorAbout 2 min
The acceleration is $2.0\,\mathrm{m\,s^{-2}}$ from $t=0$ to $3.0\,\mathrm{s}$, then rises linearly to $5.0\,\mathrm{m\,s^{-2}}$ at $t=7.0\,\mathrm{s}$, as shown. What is the change in velocity?
Show complete worked solution
Correct answer: C
Change in velocity is the area under the acceleration-time graph.
From 0 to 3 s: $$\Delta v_1=(2.0)(3.0)=6.0\,\mathrm{m\,s^{-1}}$$
From 3 to 7 s, use the trapezium area: $$\Delta v_2=\frac{2.0+5.0}{2}(4.0)=14\,\mathrm{m\,s^{-1}}$$
$$\Delta v=6.0+14=20\,\mathrm{m\,s^{-1}}$$ Option C is correct.
QUESTION 29[1 mark]
Multiple choiceFoundationCalculatorAbout 2 min
Which statement best describes the constant-acceleration equations used in mechanics?
Show complete worked solution
Correct answer: B
The equations are derived by applying the definitions of velocity and acceleration under the condition that acceleration is constant.
Their usefulness is supported by experimental observations of motion.
They do not apply unchanged to arbitrary variable acceleration. Therefore option B is correct.
QUESTION 30[1 mark]
Multiple choiceStandardCalculatorAbout 2 min
An electric motorcycle accelerates uniformly from rest to $72\,\mathrm{km\,h^{-1}}$ in $4.0\,\mathrm{s}$. Approximately what is its acceleration in units of $g$?
Show complete worked solution
Correct answer: C
Convert speed to SI units. $$72\,\mathrm{km\,h^{-1}}=\frac{72}{3.6}=20\,\mathrm{m\,s^{-1}}$$
In units of $g$, $$\frac{a}{g}=\frac{5.0}{9.8}=0.51\approx0.50$$ Option C is correct.
QUESTION 31[1 mark]
Multiple choiceAppliedCalculatorAbout 2 min
The kinetic energy of an object is calculated from $E_k=p^2/(2m)$. The percentage uncertainties in momentum and mass are $2\%$ and $5\%$, respectively. What is the percentage uncertainty in $E_k$?
Show complete worked solution
Correct answer: C
For $p^2$, double the percentage uncertainty: $2(2\%)=4\%$.
Division by $m$ adds its percentage uncertainty. $$\Delta E_k\%=4\%+5\%=9\%$$ Therefore option C is correct.
QUESTION 32[1 mark]
Multiple choiceFoundationCalculatorAbout 2 min
Which statement defines one newton?
Show complete worked solution
Correct answer: A
Newton’s second law gives $F=ma$.
Setting $m=1\,\mathrm{kg}$ and $a=1\,\mathrm{m\,s^{-2}}$ gives $F=1\,\mathrm{N}$.
The acceleration is in the direction of the resultant force. Option A is correct.
QUESTION 33[1 mark]
Multiple choiceAppliedCalculatorAbout 2 min
A rectangle measures $(40.0\pm0.4)\,\mathrm{cm}$ by $(25.0\pm0.3)\,\mathrm{cm}$. What is the percentage uncertainty in its perimeter?
$$\frac{1.4}{130.0}\times100\%=1.08\%\approx1.1\%$$ Option B is correct.
QUESTION 34[1 mark]
Multiple choiceAppliedNo calculatorAbout 2 min
A thermometer has reading uncertainty $\pm0.4^{\circ}\mathrm C$. It records a liquid cooling from $75.0^{\circ}\mathrm C$ to $55.0^{\circ}\mathrm C$. What is the percentage uncertainty in the temperature change?
Show complete worked solution
Correct answer: B
The measured change is $75.0-55.0=20.0^{\circ}\mathrm C$.
For a difference, add absolute uncertainties: $0.4+0.4=0.8^{\circ}\mathrm C$.
$$\frac{0.8}{20.0}\times100\%=4\%$$ so option B is correct.
QUESTION 35[1 mark]
Multiple choiceAppliedNo calculatorAbout 2 min
A ball is projected horizontally at $2.0\,\mathrm{m\,s^{-1}}$ from a cliff and reaches the ground $3.0\,\mathrm s$ later. The cliff is $8.0\,\mathrm m$ high. What is the magnitude of the ball’s displacement?
Show complete worked solution
Correct answer: C
Horizontal displacement is $x=v_xt=(2.0)(3.0)=6.0\,\mathrm m$.
Vertical displacement magnitude is the cliff height, $y=8.0\,\mathrm m$.
$$s=\sqrt{x^2+y^2}=\sqrt{6.0^2+8.0^2}=10.0\,\mathrm m$$ so option C is correct.
QUESTION 36[1 mark]
Multiple choiceAppliedCalculatorAbout 2 min
A ball has mass $(50\pm1)\,\mathrm g$ and speed $(25\pm1)\,\mathrm{m\,s^{-1}}$. What is the fractional uncertainty in its momentum?
Show complete worked solution
Correct answer: C
Momentum is the product $p=mv$, so fractional uncertainties add.
Mass fractional uncertainty is $1/50=0.02$ and speed fractional uncertainty is $1/25=0.04$.
The total is $0.02+0.04=0.06$, so C is correct.
QUESTION 37[1 mark]
Multiple choiceStandardNo calculatorAbout 2 min
Two fitted lines have equal absolute uncertainty in their y-intercepts. The square-data intercept is half the circle-data intercept. Which comparison is correct?
Show complete worked solution
Correct answer: C
The problem states that absolute uncertainties are equal.
Fractional uncertainty is absolute uncertainty divided by measured value.
With half the intercept but the same absolute uncertainty, the square data have twice the fractional uncertainty, so C is correct.
QUESTION 38[1 mark]
Multiple choiceFoundationNo calculatorAbout 2 min
A stone falls through air from a tall building and is still below terminal speed. How does its speed change?
Show complete worked solution
Correct answer: D
Weight initially makes the stone accelerate downward.
As speed rises, air resistance rises and reduces the resultant downward force.
Speed still increases, but acceleration and therefore its rate of increase become smaller; D is correct.
QUESTION 39[1 mark]
Multiple choiceAppliedCalculatorAbout 3 min
An object has position $x=t^2$ metres from $t=0$ to $4.0\,\mathrm s$. At what time does its instantaneous speed equal its average speed over the interval?
Show complete worked solution
Correct answer: B
Average speed is total displacement divided by time: $(16-0)/4=4\,\mathrm{m\,s^{-1}}$.
Instantaneous speed is $v=dx/dt=2t$.
Set $2t=4$ to obtain $t=2.0\,\mathrm s$, so B is correct.
QUESTION 40[1 mark]
Multiple choiceFoundationNo calculatorAbout 2 min
Which quantity has the same SI unit as energy stored per unit volume?
Show complete worked solution
Correct answer: D
Energy per volume has unit $\mathrm{J\,m^{-3}}$.
$1\,\mathrm J=1\,\mathrm{N\,m}$, so this becomes $\mathrm{N\,m^{-2}}$.
A newton per square metre is a pascal, the unit of pressure, so D is correct.
QUESTION 41[1 mark]
Multiple choiceFoundationNo calculatorAbout 2 min
Which quantity has fundamental SI units $\mathrm{kg\,m^{-1}\,s^{-2}}$?
Show complete worked solution
Correct answer: D
Pressure unit is pascal, $\mathrm{Pa=N\,m^{-2}}$.
Newton has units $\mathrm{kg\,m\,s^{-2}}$.
Dividing by $m^2$ gives $\mathrm{kg\,m^{-1}\,s^{-2}}$, so D is correct.
QUESTION 42[1 mark]
Multiple choiceAppliedCalculatorAbout 3 min
A ball falls from rest with $g=10\,\mathrm{m\,s^{-2}}$. What distances does it travel during the first, second and third one-second intervals?
Show complete worked solution
Correct answer: B
Total distance after t seconds is $s=\tfrac12gt^2=5t^2$.
Values at 0,1,2,3 s are 0,5,20,45 m.
Successive differences are 5,15,25 m, so B is correct.
QUESTION 43[1 mark]
Multiple choiceFoundationNo calculatorAbout 2 min
The constant-acceleration equations used in mechanics were established primarily through which process?
Show complete worked solution
Correct answer: A
The equations describe a model in which acceleration is constant.
Their validity is tested by measuring displacement, velocity and time in real motion.
They were developed and supported through observation and experiment. Therefore option A is correct.
QUESTION 44[1 mark]
Multiple choiceAppliedCalculatorAbout 3 min
A derived quantity is $x=A^2B/C$. Percentage uncertainties in A, B and C are 4%, 3% and 5%. What is the percentage uncertainty in x?
Show complete worked solution
Correct answer: C
For powers and products, percentage uncertainties add with the magnitude of each power.
Thus $\Delta x/x=2(4\%)+3\%+5\%$.
The total percentage uncertainty is 16%. Therefore option C is correct.
QUESTION 45[1 mark]
Multiple choiceStandardNo calculatorAbout 2 min
A model uses $p=x+yT$, where p is in pascals and T in kelvin. What are the SI units of x and y?
Show complete worked solution
Correct answer: A
Quantities added in an equation must have identical units, so x has the unit of p.
The term $yT$ must also have unit Pa, giving $[y]=\mathrm{Pa/K}$.
Thus x is in Pa and y in $\mathrm{Pa\,K^{-1}}$. Therefore option A is correct.
QUESTION 46[1 mark]
Multiple choiceAppliedCalculatorAbout 3 min
A car changes speed uniformly from rest to $108\,\mathrm{km\,h^{-1}}$ in $4.0\,\mathrm s$. What is its acceleration in units of g, using $g=9.8\,\mathrm{m\,s^{-2}}$?
$a/g=7.5/9.8=0.77$. Therefore option C is correct.
QUESTION 47[1 mark]
Multiple choiceAppliedCalculatorAbout 3 min
A projectile is launched horizontally. After $0.40\,\mathrm s$ it has travelled horizontal distance x and fallen distance y. What are its launch speed and fall distance after $0.80\,\mathrm s$?
Show complete worked solution
Correct answer: D
Horizontal speed is constant, so $u=x/0.40$.
Vertical fall from rest is $y=\tfrac12gt^2$, so doubling time multiplies distance by four.
The values are $x/0.40$ and $4y$. Therefore option D is correct.
QUESTION 48[1 mark]
Multiple choiceStandardNo calculatorAbout 2 min
A circular plate has measured radius R with absolute uncertainty $\Delta R$. What is the approximate fractional uncertainty in its area?
Show complete worked solution
Correct answer: D
Area is $A=\pi R^2$.
For a power law, fractional uncertainty is multiplied by the power.
Thus $\Delta A/A\approx2\Delta R/R$. Therefore option D is correct.
QUESTION 49[1 mark]
Multiple choiceAppliedCalculatorAbout 3 min
A ball is projected horizontally at $18\,\mathrm{m\,s^{-1}}$ from a cliff $45\,\mathrm m$ high. Neglect air resistance and use $g=10\,\mathrm{m\,s^{-2}}$. How far from the cliff does it land?
Show complete worked solution
Correct answer: B
Vertical motion gives $45=\tfrac12(10)t^2$, so $t=3.0\,\mathrm s$.
Horizontal velocity remains 18 m s$^{-1}$.
Horizontal distance is $x=18(3.0)=54\,\mathrm m$. Therefore option B is correct.
QUESTION 50[1 mark]
Multiple choiceAppliedCalculatorAbout 3 min
A cube has side length with 3% uncertainty and mass with 5% uncertainty. What is the approximate percentage uncertainty in density $\rho=m/L^3$?
Show complete worked solution
Correct answer: D
Fractional uncertainties add for products and powers.
$\Delta\rho/\rho=5\%+3(3\%)=14\%$.
The result is 14% Therefore option D is correct.
QUESTION 51[1 mark]
Multiple choiceAppliedCalculatorAbout 3 min
A vehicle moving at $24\,\mathrm{m\,s^{-1}}$ brakes uniformly at $6.0\,\mathrm{m\,s^{-2}}$. What stopping distance is required?
Show complete worked solution
Correct answer: A
Use $v^2=u^2+2as$ with final speed zero.
$0=24^2-2(6.0)s$, so $s=576/12=48\,\mathrm m$.
The result is 48 m Therefore option A is correct.
QUESTION 52[1 mark]
Multiple choiceStandardNo calculatorAbout 2 min
A runner starts from rest and accelerates constantly. Which relation describes speed v against distance s?
Show complete worked solution
Correct answer: B
For constant acceleration from rest, $v^2=2as$.
Taking the positive square root gives $v=\sqrt{2a}\sqrt{s}$.
The result is $v\propto\sqrt{s}$ Therefore option B is correct.
QUESTION 53[1 mark]
Multiple choiceStandardNo calculatorAbout 2 min
A projectile moves from launch to its highest point with air resistance negligible. What happens to the magnitudes of its velocity components?
Show complete worked solution
Correct answer: C
Gravity has no horizontal component, so horizontal velocity is constant.
The downward acceleration reduces the upward vertical component to zero at the top.
The result is Horizontal stays constant; vertical decreases. Therefore option C is correct.
QUESTION 54[1 mark]
Multiple choiceFoundationNo calculatorAbout 2 min
An object completes 7 revolutions in 4.0 s. What is its frequency?
Show complete worked solution
Correct answer: B
Frequency is the number of cycles per unit time.
$f=N/t=7/4.0=1.75\,\mathrm{Hz}$.
The result is 1.75 Hz Therefore option B is correct.
QUESTION 55[1 mark]
Multiple choiceStandardNo calculatorAbout 2 min
The measured diameter of a spherical sample is D with absolute uncertainty $\Delta D$. Estimate the fractional uncertainty in its calculated volume.
Show complete worked solution
Correct answer: D
The sphere volume is proportional to the cube of its diameter.
For $V\propto D^3$, fractional uncertainty is approximately three times the fractional diameter uncertainty.
The result is $3\Delta D/D$ Therefore option D is correct.
QUESTION 56[1 mark]
Multiple choiceAppliedCalculatorAbout 3 min
A ball projected vertically passes the same height at times 3.0 s and 9.0 s. Use $g=10\,\mathrm{m\,s^{-2}}$. What was its launch speed?
Show complete worked solution
Correct answer: A
For vertical projectile motion, the two times at one height are symmetric about time to maximum height.
$t_{top}=(3.0+9.0)/2=6.0\,\mathrm s$, so $u=gt_{top}=60\,\mathrm{m\,s^{-1}}$.
The result is 60 m s$^{-1}$ Therefore option A is correct.
QUESTION 57[1 mark]
Multiple choiceAppliedCalculatorAbout 3 min
A particle starts from rest and moves with constant acceleration through $0.072\,\mathrm m$ in $6.0\,\mathrm{ms}$. What is its acceleration?
The result is $4.0\times10^3\,\mathrm{m\,s^{-2}}$ Therefore option A is correct.
QUESTION 58[1 mark]
Multiple choiceStandardNo calculatorAbout 2 min
When checking a derived kinematics result from a satellite measurement using a reference sample, which statement about a limiting case is valid?
Show complete worked solution
Correct answer: B
Inspect $v^2=u^2+2as$ as displacement becomes very small or very large.
Compare the mathematical trend with the physical meaning of the variables and the domain of the model.
The supported conclusion is speed squared follows the limiting behaviour implied by $v^2=u^2+2as$. This corresponds to option B.
QUESTION 59[1 mark]
Multiple choiceAppliedCalculatorAbout 2 min
During a wave-tank investigation using a reference sample, displacement increases by $70\%$. What percentage change in speed squared follows from $v^2=u^2+2as$?
Show complete worked solution
Correct answer: C
Convert the percentage change to the multiplier $x_2/x_1=1.7$.
Use $y_2/y_1=(1.7)^{1}=1.7$ and convert the ratio back to a percentage change.
The supported conclusion is $70\%$ increase This corresponds to option C.
QUESTION 60[1 mark]
Multiple choiceAppliedCalculatorAbout 2 min
In a particle-detector experiment with negligible losses, displacement is multiplied by $1.9$ and then by $0.77$. By what overall factor does speed squared change according to $v^2=u^2+2as$?
Show complete worked solution
Correct answer: D
The total multiplier of displacement is $1.9\times0.77=1.463$.
Raise this multiplier to the power $1$, giving $1.463$.
The supported conclusion is $1.463$ This corresponds to option D.
QUESTION 61[1 mark]
Multiple choiceAdvancedCalculatorAbout 2 min
In an optical-instrument test using two matched systems, a student says the uncertainty result is exact because it matches $\frac{\Delta Q}{Q}=n\frac{\Delta x}{x}$. Which evaluation is strongest?
Show complete worked solution
Correct answer: A
Separate the mathematical model from the measurement used to test it.
State model assumptions, measurement uncertainty and the tested range before judging agreement.
The supported conclusion is The claim is valid only under the assumptions required by $\frac{\Delta Q}{Q}=n\frac{\Delta x}{x}$. This corresponds to option A.
QUESTION 62[1 mark]
Multiple choiceAdvancedCalculatorAbout 2 min
Data from an optical-instrument test over a controlled range agree with the kinematics prediction within uncertainty. Which conclusion is justified?
Show complete worked solution
Correct answer: B
Experimental agreement is always limited by the range and quality of the measurements.
State support within uncertainty without extending the conclusion beyond the tested conditions.
The supported conclusion is The measured trend supports $v^2=u^2+2as$ within uncertainty; it does not prove the model outside the tested range. This corresponds to option B.
QUESTION 63[1 mark]
Multiple choiceAppliedCalculatorAbout 2 min
In an astronomical observation with uncertainty bars, two kinematics systems have displacement values in the ratio $1.75:1.42$. What is the ratio of their speed squared values when the other terms in $v^2=u^2+2as$ are equal?
Show complete worked solution
Correct answer: B
Write $y_1/y_2=(x_1/x_2)^{1}$.
Substitute the stated ratio: $(1.75/1.42)^{1}=1.232$.
The supported conclusion is $1.232$ This corresponds to option B.
QUESTION 64[1 mark]
Multiple choiceAdvancedCalculatorAbout 2 min
In a thermal-control system with fixed geometry, a student says the kinematics result is exact because it matches $v^2=u^2+2as$. Which evaluation is strongest?
Show complete worked solution
Correct answer: A
Separate the mathematical model from the measurement used to test it.
State model assumptions, measurement uncertainty and the tested range before judging agreement.
The supported conclusion is The claim is valid only under the assumptions required by $v^2=u^2+2as$. This corresponds to option A.
QUESTION 65[1 mark]
Multiple choiceAppliedCalculatorAbout 3 min
Which graph would best test the predicted dependence in an environmental monitor using a digital sensor involving kinematics?
Show complete worked solution
Correct answer: D
Rearrange $v^2=u^2+2as$ into a linear form.
Choose axes that predict a straight line, then assess gradient, intercept and uncertainty bars.
The supported conclusion is Plot speed squared against a transformed displacement chosen to make $v^2=u^2+2as$ linear. This corresponds to option D.
QUESTION 66[1 mark]
Multiple choiceAdvancedCalculatorAbout 2 min
In a thermal-control system under steady conditions, a student says the kinematics result is exact because it matches $v^2=u^2+2as$. Which evaluation is strongest?
Show complete worked solution
Correct answer: A
Separate the mathematical model from the measurement used to test it.
State model assumptions, measurement uncertainty and the tested range before judging agreement.
The supported conclusion is The claim is valid only under the assumptions required by $v^2=u^2+2as$. This corresponds to option A.
QUESTION 67[1 mark]
Multiple choiceAdvancedCalculatorAbout 2 min
Data from an optical-instrument test under steady conditions agree with the kinematics prediction within uncertainty. Which conclusion is justified?
Show complete worked solution
Correct answer: B
Experimental agreement is always limited by the range and quality of the measurements.
State support within uncertainty without extending the conclusion beyond the tested conditions.
The supported conclusion is The measured trend supports $v^2=u^2+2as$ within uncertainty; it does not prove the model outside the tested range. This corresponds to option B.
QUESTION 68[1 mark]
Multiple choiceAdvancedCalculatorAbout 3 min
In a circuit-design trial with negligible losses, by what factor must displacement change for speed squared to change by a factor of $2.25$?
Show complete worked solution
Correct answer: C
Use the ratio equation $2.25=(x_2/x_1)^{1}$.
Take the power $1/1$ to obtain $x_2/x_1=2.25$.
The supported conclusion is $2.25$ This corresponds to option C.
QUESTION 69[1 mark]
Multiple choiceStandardNo calculatorAbout 2 min
Which condition is required when using $v^2=u^2+2as$ to compare two measurements of speed squared in a nuclear-monitoring station with negligible losses?
Show complete worked solution
Correct answer: A
Identify which variables appear in the complete physical relation.
A one-variable proportional comparison is valid only when the remaining variables are controlled.
The supported conclusion is The prediction follows only if the quantities omitted from $v^2=u^2+2as$ remain constant. This corresponds to option A.
QUESTION 70[1 mark]
Multiple choiceAdvancedCalculatorAbout 2 min
After calculating speed squared for a renewable-energy trial using a digital sensor, which combined check is most useful?
Show complete worked solution
Correct answer: D
Test the numerical answer independently of the arithmetic used to obtain it.
Check units, sign or direction, order of magnitude and a physically sensible limiting case.
The supported conclusion is The result should agree with $v^2=u^2+2as$ in sign, magnitude, dimensions and limiting behaviour. This corresponds to option D.
QUESTION 71[1 mark]
Multiple choiceAdvancedCalculatorAbout 2 min
For a particle-detector experiment using a calibrated scale, a log–log graph of speed squared against displacement tests the dependence represented by $v^2=u^2+2as$. What gradient is predicted?
Show complete worked solution
Correct answer: C
For a power law $y=Cx^n$, take logarithms: $\log y=\log C+n\log x$.
The gradient equals the exponent, so here it is $1$.
The supported conclusion is $1$ This corresponds to option C.
QUESTION 72[1 mark]
Multiple choiceStandardNo calculatorAbout 2 min
A graph from an optical-instrument test with repeated readings shows fractional uncertainty varying with input uncertainty in the trend illustrated. Which conclusion is justified?
Show complete worked solution
Correct answer: D
Read the axes and overall trend before using any individual point.
Compare the trend with $\frac{\Delta Q}{Q}=n\frac{\Delta x}{x}$ while holding the other variables constant.
The supported conclusion is fractional uncertainty changes consistently with $\frac{\Delta Q}{Q}=n\frac{\Delta x}{x}$. This corresponds to option D.
QUESTION 73[1 mark]
Multiple choiceStandardNo calculatorAbout 2 min
When checking a derived kinematics result from an optical-instrument test with repeated readings, which statement about a limiting case is valid?
Show complete worked solution
Correct answer: A
Inspect $v^2=u^2+2as$ as displacement becomes very small or very large.
Compare the mathematical trend with the physical meaning of the variables and the domain of the model.
The supported conclusion is speed squared follows the limiting behaviour implied by $v^2=u^2+2as$. This corresponds to option A.
QUESTION 74[1 mark]
Multiple choiceAppliedCalculatorAbout 2 min
In a field-mapping exercise with repeated readings, displacement is multiplied by $1.85$ and then by $0.77$. By what overall factor does speed squared change according to $v^2=u^2+2as$?
Show complete worked solution
Correct answer: C
The total multiplier of displacement is $1.85\times0.77=1.425$.
Raise this multiplier to the power $1$, giving $1.425$.
The supported conclusion is $1.425$ This corresponds to option C.
QUESTION 75[1 mark]
Multiple choiceAdvancedCalculatorAbout 3 min
In a spacecraft instrument under steady conditions, by what factor must displacement change for speed squared to change by a factor of $2.75$?
Show complete worked solution
Correct answer: B
Use the ratio equation $2.75=(x_2/x_1)^{1}$.
Take the power $1/1$ to obtain $x_2/x_1=2.75$.
The supported conclusion is $2.75$ This corresponds to option B.
QUESTION 76[1 mark]
Multiple choiceStandardNo calculatorAbout 2 min
When checking a derived kinematics result from an optical-instrument test using a computer interface, which statement about a limiting case is valid?
Show complete worked solution
Correct answer: A
Inspect $v^2=u^2+2as$ as displacement becomes very small or very large.
Compare the mathematical trend with the physical meaning of the variables and the domain of the model.
The supported conclusion is speed squared follows the limiting behaviour implied by $v^2=u^2+2as$. This corresponds to option A.
QUESTION 77[1 mark]
Multiple choiceStandardNo calculatorAbout 2 min
For an engineering prototype using a reference sample, with the other quantities fixed, which proportionality between speed squared and displacement is consistent with $v^2=u^2+2as$?
Show complete worked solution
Correct answer: D
Collect all fixed quantities into a constant $C$.
The relation reduces to $y=Cx^{1}$.
The supported conclusion is $y\propto x^{1}$ This corresponds to option D.
QUESTION 78[1 mark]
Multiple choiceStandardNo calculatorAbout 2 min
When checking a derived kinematics result from an optical-instrument test using a digital sensor, which statement about a limiting case is valid?
Show complete worked solution
Correct answer: A
Inspect $v^2=u^2+2as$ as displacement becomes very small or very large.
Compare the mathematical trend with the physical meaning of the variables and the domain of the model.
The supported conclusion is speed squared follows the limiting behaviour implied by $v^2=u^2+2as$. This corresponds to option A.
QUESTION 79[1 mark]
Multiple choiceAppliedCalculatorAbout 2 min
During a circuit-design trial using a digital sensor, displacement increases by $25\%$. What percentage change in speed squared follows from $v^2=u^2+2as$?
Show complete worked solution
Correct answer: B
Convert the percentage change to the multiplier $x_2/x_1=1.25$.
Use $y_2/y_1=(1.25)^{1}=1.25$ and convert the ratio back to a percentage change.
The supported conclusion is $25\%$ increase This corresponds to option B.
QUESTION 80[1 mark]
Multiple choiceAppliedCalculatorAbout 2 min
In a field-mapping exercise using a calibrated scale, displacement is multiplied by $1.3$ and then by $0.65$. By what overall factor does speed squared change according to $v^2=u^2+2as$?
Show complete worked solution
Correct answer: C
The total multiplier of displacement is $1.3\times0.65=0.845$.
Raise this multiplier to the power $1$, giving $0.845$.
The supported conclusion is $0.845$ This corresponds to option C.
QUESTION 81[1 mark]
Multiple choiceAdvancedCalculatorAbout 2 min
Data from a laboratory calibration using a calibrated scale agree with the kinematics prediction within uncertainty. Which conclusion is justified?
Show complete worked solution
Correct answer: A
Experimental agreement is always limited by the range and quality of the measurements.
State support within uncertainty without extending the conclusion beyond the tested conditions.
The supported conclusion is The measured trend supports $v^2=u^2+2as$ within uncertainty; it does not prove the model outside the tested range. This corresponds to option A.
QUESTION 82[1 mark]
Multiple choiceStandardNo calculatorAbout 2 min
Which relationship is the appropriate starting point for a kinematics calculation in a wave-tank investigation using two matched systems?
Show complete worked solution
Correct answer: B
List the variables described by the kinematics situation.
The current-syllabus relation connecting them is $v^2=u^2+2as$.
The supported conclusion is $v^2=u^2+2as$ This corresponds to option B.
QUESTION 83[1 mark]
Multiple choiceStandardNo calculatorAbout 2 min
A graph from a thermal-control system using two matched systems shows speed squared varying with displacement in the trend illustrated. Which conclusion is justified?
Show complete worked solution
Correct answer: D
Read the axes and overall trend before using any individual point.
Compare the trend with $v^2=u^2+2as$ while holding the other variables constant.
The supported conclusion is speed squared changes consistently with $v^2=u^2+2as$. This corresponds to option D.
QUESTION 84[1 mark]
Multiple choiceAdvancedCalculatorAbout 2 min
For a particle-detector experiment at low amplitude, a log–log graph of speed squared against displacement tests the dependence represented by $v^2=u^2+2as$. What gradient is predicted?
Show complete worked solution
Correct answer: C
For a power law $y=Cx^n$, take logarithms: $\log y=\log C+n\log x$.
The gradient equals the exponent, so here it is $1$.
The supported conclusion is $1$ This corresponds to option C.
QUESTION 85[1 mark]
Multiple choiceStandardNo calculatorAbout 2 min
A graph from an optical-instrument test with an idealized component shows fractional uncertainty varying with input uncertainty in the trend illustrated. Which conclusion is justified?
Show complete worked solution
Correct answer: D
Read the axes and overall trend before using any individual point.
Compare the trend with $\frac{\Delta Q}{Q}=n\frac{\Delta x}{x}$ while holding the other variables constant.
The supported conclusion is fractional uncertainty changes consistently with $\frac{\Delta Q}{Q}=n\frac{\Delta x}{x}$. This corresponds to option D.
QUESTION 86[1 mark]
Multiple choiceStandardNo calculatorAbout 2 min
When checking a derived kinematics result from an optical-instrument test at low amplitude, which statement about a limiting case is valid?
Show complete worked solution
Correct answer: A
Inspect $v^2=u^2+2as$ as displacement becomes very small or very large.
Compare the mathematical trend with the physical meaning of the variables and the domain of the model.
The supported conclusion is speed squared follows the limiting behaviour implied by $v^2=u^2+2as$. This corresponds to option A.
QUESTION 87[1 mark]
Multiple choiceAdvancedCalculatorAbout 2 min
In a nuclear-monitoring station with an idealized component, a student says the kinematics result is exact because it matches $v^2=u^2+2as$. Which evaluation is strongest?
Show complete worked solution
Correct answer: D
Separate the mathematical model from the measurement used to test it.
State model assumptions, measurement uncertainty and the tested range before judging agreement.
The supported conclusion is The claim is valid only under the assumptions required by $v^2=u^2+2as$. This corresponds to option D.
QUESTION 88[1 mark]
Multiple choiceStandardNo calculatorAbout 2 min
Which relationship is the appropriate starting point for a kinematics calculation in a wave-tank investigation with fixed geometry?
Show complete worked solution
Correct answer: B
List the variables described by the kinematics situation.
The current-syllabus relation connecting them is $v^2=u^2+2as$.
The supported conclusion is $v^2=u^2+2as$ This corresponds to option B.
QUESTION 89[1 mark]
Multiple choiceStandardNo calculatorAbout 2 min
When checking a derived kinematics result from an optical-instrument test using a reference sample, which statement about a limiting case is valid?
Show complete worked solution
Correct answer: A
Inspect $v^2=u^2+2as$ as displacement becomes very small or very large.
Compare the mathematical trend with the physical meaning of the variables and the domain of the model.
The supported conclusion is speed squared follows the limiting behaviour implied by $v^2=u^2+2as$. This corresponds to option A.
QUESTION 90[1 mark]
Multiple choiceAdvancedCalculatorAbout 2 min
In a nuclear-monitoring station at low amplitude, a student says the kinematics result is exact because it matches $v^2=u^2+2as$. Which evaluation is strongest?
Show complete worked solution
Correct answer: D
Separate the mathematical model from the measurement used to test it.
State model assumptions, measurement uncertainty and the tested range before judging agreement.
The supported conclusion is The claim is valid only under the assumptions required by $v^2=u^2+2as$. This corresponds to option D.
QUESTION 91[1 mark]
Multiple choiceAppliedCalculatorAbout 3 min
Which graph would best test the predicted dependence in a transport-safety test using a calibrated scale involving kinematics?
Show complete worked solution
Correct answer: C
Rearrange $v^2=u^2+2as$ into a linear form.
Choose axes that predict a straight line, then assess gradient, intercept and uncertainty bars.
The supported conclusion is Plot speed squared against a transformed displacement chosen to make $v^2=u^2+2as$ linear. This corresponds to option C.
QUESTION 92[1 mark]
Multiple choiceAppliedCalculatorAbout 3 min
In an environmental monitor with SI data, displacement is multiplied by $1.6$. What is the new-to-old ratio of speed squared predicted by $v^2=u^2+2as$?
Show complete worked solution
Correct answer: C
Write the ratio form $y_2/y_1=(x_2/x_1)^{1}$ for the stated dependence.
Substitute $x_2/x_1=1.6$ and evaluate the relevant reciprocal or power to obtain 1.6.
The supported conclusion is $1.6$ This corresponds to option C.
QUESTION 93[1 mark]
Multiple choiceStandardNo calculatorAbout 2 min
For an engineering prototype with SI data, with the other quantities fixed, which proportionality between speed squared and displacement is consistent with $v^2=u^2+2as$?
Show complete worked solution
Correct answer: D
Collect all fixed quantities into a constant $C$.
The relation reduces to $y=Cx^{1}$.
The supported conclusion is $y\propto x^{1}$ This corresponds to option D.
QUESTION 94[1 mark]
Multiple choiceAdvancedCalculatorAbout 2 min
After calculating speed squared for a renewable-energy trial using a computer interface, which combined check is most useful?
Show complete worked solution
Correct answer: D
Test the numerical answer independently of the arithmetic used to obtain it.
Check units, sign or direction, order of magnitude and a physically sensible limiting case.
The supported conclusion is The result should agree with $v^2=u^2+2as$ in sign, magnitude, dimensions and limiting behaviour. This corresponds to option D.
QUESTION 95[1 mark]
Multiple choiceStandardNo calculatorAbout 2 min
Which relationship is the appropriate starting point for a kinematics calculation in a wave-tank investigation with uncertainty bars?
Show complete worked solution
Correct answer: B
List the variables described by the kinematics situation.
The current-syllabus relation connecting them is $v^2=u^2+2as$.
The supported conclusion is $v^2=u^2+2as$ This corresponds to option B.
QUESTION 96[1 mark]
Multiple choiceAdvancedCalculatorAbout 2 min
For a particle-detector experiment with uncertainty bars, a log–log graph of speed squared against displacement tests the dependence represented by $v^2=u^2+2as$. What gradient is predicted?
Show complete worked solution
Correct answer: C
For a power law $y=Cx^n$, take logarithms: $\log y=\log C+n\log x$.
The gradient equals the exponent, so here it is $1$.
The supported conclusion is $1$ This corresponds to option C.
QUESTION 97[1 mark]
Multiple choiceAdvancedCalculatorAbout 2 min
Data from a laboratory calibration under steady conditions agree with the kinematics prediction within uncertainty. Which conclusion is justified?
Show complete worked solution
Correct answer: A
Experimental agreement is always limited by the range and quality of the measurements.
State support within uncertainty without extending the conclusion beyond the tested conditions.
The supported conclusion is The measured trend supports $v^2=u^2+2as$ within uncertainty; it does not prove the model outside the tested range. This corresponds to option A.
QUESTION 98[1 mark]
Multiple choiceAppliedCalculatorAbout 2 min
In a satellite measurement with SI data, two kinematics systems have displacement values in the ratio $1.35:1.42$. What is the ratio of their speed squared values when the other terms in $v^2=u^2+2as$ are equal?
Show complete worked solution
Correct answer: A
Write $y_1/y_2=(x_1/x_2)^{1}$.
Substitute the stated ratio: $(1.35/1.42)^{1}=0.951$.
The supported conclusion is $0.951$ This corresponds to option A.
QUESTION 99[1 mark]
Multiple choiceStandardNo calculatorAbout 2 min
A graph from a thermal-control system using a computer interface shows speed squared varying with displacement in the trend illustrated. Which conclusion is justified?
Show complete worked solution
Correct answer: D
Read the axes and overall trend before using any individual point.
Compare the trend with $v^2=u^2+2as$ while holding the other variables constant.
The supported conclusion is speed squared changes consistently with $v^2=u^2+2as$. This corresponds to option D.
QUESTION 100[1 mark]
Multiple choiceStandardNo calculatorAbout 2 min
When checking a derived kinematics result from an optical-instrument test using a computer interface, which statement about a limiting case is valid?
Show complete worked solution
Correct answer: A
Inspect $v^2=u^2+2as$ as displacement becomes very small or very large.
Compare the mathematical trend with the physical meaning of the variables and the domain of the model.
The supported conclusion is speed squared follows the limiting behaviour implied by $v^2=u^2+2as$. This corresponds to option A.
QUESTION 101[1 mark]
Multiple choiceAdvancedCalculatorAbout 2 min
For a particle-detector experiment with an idealized component, a log–log graph of speed squared against displacement tests the dependence represented by $v^2=u^2+2as$. What gradient is predicted?
Show complete worked solution
Correct answer: C
For a power law $y=Cx^n$, take logarithms: $\log y=\log C+n\log x$.
The gradient equals the exponent, so here it is $1$.
The supported conclusion is $1$ This corresponds to option C.
QUESTION 102[1 mark]
Multiple choiceAdvancedCalculatorAbout 2 min
Data from a laboratory calibration using a calibrated scale agree with the kinematics prediction within uncertainty. Which conclusion is justified?
Show complete worked solution
Correct answer: A
Experimental agreement is always limited by the range and quality of the measurements.
State support within uncertainty without extending the conclusion beyond the tested conditions.
The supported conclusion is The measured trend supports $v^2=u^2+2as$ within uncertainty; it does not prove the model outside the tested range. This corresponds to option A.
QUESTION 103[1 mark]
Multiple choiceAdvancedCalculatorAbout 2 min
After calculating speed squared for a renewable-energy trial over a controlled range, which combined check is most useful?
Show complete worked solution
Correct answer: D
Test the numerical answer independently of the arithmetic used to obtain it.
Check units, sign or direction, order of magnitude and a physically sensible limiting case.
The supported conclusion is The result should agree with $v^2=u^2+2as$ in sign, magnitude, dimensions and limiting behaviour. This corresponds to option D.
QUESTION 104[1 mark]
Multiple choiceAppliedCalculatorAbout 2 min
In a satellite measurement over a controlled range, two kinematics systems have displacement values in the ratio $1.65:1.1$. What is the ratio of their speed squared values when the other terms in $v^2=u^2+2as$ are equal?
Show complete worked solution
Correct answer: A
Write $y_1/y_2=(x_1/x_2)^{1}$.
Substitute the stated ratio: $(1.65/1.1)^{1}=1.5$.
The supported conclusion is $1.5$ This corresponds to option A.
QUESTION 105[9 marks]
Multi-partAppliedCalculatorAbout 11.2 min
A research satellite follows a circular orbit of radius $7.20\times10^6\,\mathrm{m}$ and completes one orbit every $96.0\,\mathrm{min}$.
a. Calculate the satellite's average speed during one orbit.[3]
b. Describe the satellite's instantaneous velocity.[2]
c. Determine its displacement from its starting point after (i) $384\,\mathrm{min}$ and (ii) $432\,\mathrm{min}$.[4]
Final answer: $7.85\times10^3\,\mathrm{m\,s^{-1}}$
(b)
The speed remains constant.
At every point the velocity is tangent to the circular path.
Its direction therefore changes continuously, even though its magnitude does not.
Final answer: Constant magnitude and continuously changing direction, tangent to the orbit.
(c)
$$\frac{384}{96}=4.00\text{ orbits}$$
The satellite returns to its starting point, so
$$\boxed{\Delta r=0}$$
$$\frac{432}{96}=4.50\text{ orbits}$$
It is diametrically opposite its starting point.
$$|\Delta r|=2r=2(7.20\times10^6)=\boxed{1.44\times10^7\,\mathrm{m}}$$
Final answer: (i) $0$; (ii) $1.44\times10^7\,\mathrm{m}$ toward the opposite side of the orbit.
QUESTION 106[9 marks]
Data-basedAppliedCalculatorAbout 11.2 min
The graph shows the distance travelled by a metro train between two stations.
a. Describe the train's motion during the three straight-line intervals.[3]
b. State the distance between the stations.[1]
c. Calculate the maximum speed.[3]
d. Determine the average speed for the complete journey.[2]
Show complete worked solution
(a)
Each positive straight-line gradient represents a constant positive speed.
The middle segment is steepest, so the train is fastest there.
The final segment is less steep, so the train continues forward at a lower speed.
Final answer: It moves slowly, then faster, then more slowly; it never reverses or stops within an interval.
(b)
The final ordinate of the distance-time graph is
$$\boxed{3500\,\mathrm{m}}$$
Final answer: $3500\,\mathrm{m}$
(c)
The middle segment has the greatest gradient.
$$v_{max}=\frac{2900-500}{190-70}=\frac{2400}{120}$$
$$\boxed{v_{max}=20.0\,\mathrm{m\,s^{-1}}}$$
A survey boat points perpendicular to a river bank and moves at $5.35\,\mathrm{m\,s^{-1}}$ relative to the water. The current is $3.05\,\mathrm{m\,s^{-1}}$ downstream and the river is $116\,\mathrm{m}$ wide.
a. Calculate the crossing time.[2]
b. Determine the downstream drift.[2]
c. Determine the boat's speed and direction relative to the bank.[4]
Final answer: $6.16\,\mathrm{m\,s^{-1}}$ at $29.7^\circ$ downstream from perpendicular
QUESTION 108[15 marks]
Data-basedAdvancedCalculatorAbout 18.8 min
The velocity-time graph describes an automated shuttle during a $16.0\,\mathrm{s}$ test.
a. Describe the motion in each interval and state when the shuttle reverses direction.[4]
b. Calculate the acceleration during $0$-$4\,\mathrm{s}$ and $9$-$13\,\mathrm{s}$.[3]
c. Determine the total distance travelled and the displacement.[6]
d. Calculate the average velocity.[2]
Show complete worked solution
(a)
$0$-$4\,\mathrm{s}$: velocity rises from $0$ to $12\,\mathrm{m\,s^{-1}}$.
$4$-$9\,\mathrm{s}$: velocity is constant and positive.
$9$-$13\,\mathrm{s}$: acceleration is negative and the velocity crosses zero at $t=12\,\mathrm{s}$.
$13$-$16\,\mathrm{s}$: the shuttle moves in the negative direction while slowing to rest.
Final answer: It accelerates forward, travels forward at constant velocity, decelerates through rest at $12.0\,\mathrm{s}$, then returns and stops at $16.0\,\mathrm{s}$.
Final answer: $24.6\,\mathrm{m\,s^{-1}}$ at $53.6^\circ$ below horizontal
QUESTION 111[7 marks]
Data-basedAppliedCalculatorAbout 8.8 min
An inspection robot has the velocity-time graph shown. Positive velocity is east.
a. State the times at which the robot is stationary and identify when it reverses direction.[1]
b. Calculate the displacement during the complete motion.[4]
c. Calculate the total distance travelled.[2]
Show complete worked solution
(a)
Stationary motion corresponds to $v=0$.
The graph meets the time axis at $t=0$, $9.0$ and $20.0\,\mathrm{s}$.
The velocity changes from positive to negative at $9.0\,\mathrm{s}$, so that is the reversal time.
Final answer: $t=0$, $9.0$ and $20.0\,\mathrm{s}$; it reverses at $9.0\,\mathrm{s}$.
(b)
Eastward area:
$$A_E=\tfrac12(3)(8)+(3)(8)+\tfrac12(3)(8)=48\,\mathrm{m}$$
Westward area magnitude:
$$A_W=\tfrac12(5)(6)+\tfrac12(6)(6)=33\,\mathrm{m}$$
$$\Delta x=48-33=\boxed{15.0\,\mathrm{m}\text{ east}}$$
Final answer: $15.0\,\mathrm{m}$ east.
(c)
Distance adds the magnitudes of all areas.
$$d=48+33=\boxed{81.0\,\mathrm{m}}$$
Final answer: $81.0\,\mathrm{m}$
QUESTION 112[8 marks]
Multi-partAdvancedCalculatorAbout 10 min
A car is $120\,\mathrm{m}$ east of an observer and approaches at a constant velocity of $24.0\,\mathrm{m\,s^{-1}}$ west. It passes the observer, continues for $2.50\,\mathrm{s}$, and then brakes uniformly to rest $132\,\mathrm{m}$ west of the observer.
a. Choose east as positive and sketch the velocity-time graph from the initial instant until the car stops.[3]
b. Determine the acceleration while braking.[3]
c. Calculate the time at which the car stops, measured from the initial instant.[2]
Show complete worked solution
(a)
The car reaches the observer after
$$t=\frac{120}{24.0}=5.00\,\mathrm{s}$$
Braking begins at $5.00+2.50=7.50\,\mathrm{s}$.
The graph remains at $-24.0\,\mathrm{m\,s^{-1}}$ until $7.50\,\mathrm{s}$ and then rises linearly to zero.
Final answer: A horizontal line at $-24.0\,\mathrm{m\,s^{-1}}$ until $7.50\,\mathrm{s}$, then a straight rise to zero at $13.5\,\mathrm{s}$.
(b)
At braking start the car is $60.0\,\mathrm{m}$ west of the observer. It stops at $132\,\mathrm{m}$ west, so $s=-72.0\,\mathrm{m}$.
$$v^2=u^2+2as$$
$$0=(-24.0)^2+2a(-72.0)$$
$$a=\boxed{+4.00\,\mathrm{m\,s^{-2}}}$$
A ball is released from rest above a rigid floor. It bounces three times, reaching a smaller height after every impact. The velocity-time sketch uses upward as positive.
a. Explain the sloping sections between impacts.[2]
b. Explain the near-vertical changes in velocity at each impact.[2]
c. State why successive positive velocity peaks become smaller.[1]
d. Sketch the corresponding acceleration-time graph, ignoring the finite collision time.[3]
Show complete worked solution
(a)
Between impacts, the only significant force is weight.
Therefore
$$a=\frac{\Delta v}{\Delta t}=-g$$
and every free-flight segment has the same negative gradient.
Final answer: They have constant gradient $-g$ because gravity gives constant downward acceleration.
(b)
Immediately before impact the velocity is negative.
The floor exerts a large upward force for a short collision time, producing an upward impulse and a rapid change to positive velocity.
Final answer: The floor supplies a large upward impulse in a very short time, reversing the velocity.
(c)
Sound, thermal energy and deformation remove mechanical energy during each collision, so the rebound speed decreases.
Final answer: Mechanical energy is dissipated during each inelastic collision.
(d)
Between impacts, draw a horizontal line at $a=-g$.
At each impact, draw a very narrow positive spike because the upward contact acceleration is large and brief.
The areas of the spikes decrease as the rebound speeds decrease.
Final answer: A constant line at $-g$ with narrow positive impulse spikes at impact times.
QUESTION 114[10 marks]
Data-basedStandardCalculatorAbout 12.5 min
A ball is projected from level ground at $24.0\,\mathrm{m\,s^{-1}}$ and $45^\circ$ above the horizontal. Ignore air resistance.
a. Calculate the horizontal and vertical components of the launch velocity.[3]
b. Determine the maximum height.[3]
c. Calculate the flight time and horizontal range.[4]
One length takes
$$\frac{25.0}{1.25}=20.0\,\mathrm{s}$$
The displacement alternates between $0$ and $25.0\,\mathrm{m}$ every $20.0\,\mathrm{s}$.
The required points are $(0,0)$, $(20,25)$, $(40,0)$, $(60,25)$ and $(80,0)$, joined by straight lines.
Final answer: A triangular graph through $(0,0)$, $(20,25)$, $(40,0)$, $(60,25)$ and $(80,0)$.
(c)
The swimmer finishes at the starting point, so total displacement is zero.
$$\bar v=\frac{0}{80.0}=\boxed{0\,\mathrm{m\,s^{-1}}}$$
$$\text{average speed}=\frac{100}{80.0}=\boxed{1.25\,\mathrm{m\,s^{-1}}}$$
Final answer: $0\,\mathrm{m\,s^{-1}}$ and $1.25\,\mathrm{m\,s^{-1}}$ respectively.
QUESTION 116[8 marks]
Data-basedAppliedCalculatorAbout 10 min
A rescue package leaves a horizontal platform at $11.0\,\mathrm{m\,s^{-1}}$ from a height of $14.0\,\mathrm{m}$.
A motion logger records the velocity of a car every $2.0\,\mathrm{s}$. Assume straight-line changes between readings.
a. Plot a velocity-time graph of the readings.[2]
b. Determine the greatest average acceleration over any recorded interval.[3]
c. Estimate the distance travelled in $16.0\,\mathrm{s}$.[4]
Show complete worked solution
(a)
Plot time on the horizontal axis from $0$ to $16\,\mathrm{s}$ and velocity on the vertical axis from $0$ to $22\,\mathrm{m\,s^{-1}}$.
Plot every table pair and join consecutive points with straight lines.
Final answer: The nine points plotted with labelled axes and consecutive points joined.
(b)
The largest two-second velocity increase is from $11.9$ to $17.6\,\mathrm{m\,s^{-1}}$.
$$a=\frac{17.6-11.9}{10-8}=\frac{5.7}{2.0}=\boxed{2.85\,\mathrm{m\,s^{-2}}}$$
Final answer: $2.85\,\mathrm{m\,s^{-2}}$ from $8$ to $10\,\mathrm{s}$.
(c)
Apply the trapezoidal rule with interval width $2.0\,\mathrm{s}$.
$$s=\sum \frac{v_i+v_{i+1}}{2}(2.0)$$
$$s=1.2+4.0+9.3+18.4+29.5+37.7+41.6+43.5$$
$$s=\boxed{185.2\,\mathrm{m}\approx185\,\mathrm{m}}$$
Final answer: $185\,\mathrm{m}$ approximately.
QUESTION 119[10 marks]
Data-basedAdvancedCalculatorAbout 12.5 min
A ball is projected from level ground at $28.0\,\mathrm{m\,s^{-1}}$ and $55^\circ$ above the horizontal. Ignore air resistance.
a. Calculate the horizontal and vertical components of the launch velocity.[3]
b. Determine the maximum height.[3]
c. Calculate the flight time and horizontal range.[4]
A survey boat points perpendicular to a river bank and moves at $4.00\,\mathrm{m\,s^{-1}}$ relative to the water. The current is $2.00\,\mathrm{m\,s^{-1}}$ downstream and the river is $80\,\mathrm{m}$ wide.
a. Calculate the crossing time.[2]
b. Determine the downstream drift.[2]
c. Determine the boat's speed and direction relative to the bank.[4]
For uniform acceleration,
$$\bar v=\frac{u+v}{2}=\frac{78.0+0}{2}=39.0\,\mathrm{m\,s^{-1}}$$
$$s=\bar vt=(39.0)(72.0)=\boxed{2.81\times10^3\,\mathrm{m}}$$
Final answer: $2.81\times10^3\,\mathrm{m}$
QUESTION 130[11 marks]
Data-basedAppliedCalculatorAbout 13.8 min
The graph represents the displacement of a runner along a straight path.
a. Describe the motion from $0$ to $14\,\mathrm{s}$.[4]
b. Calculate the velocity during the first and third intervals.[4]
c. Determine the total distance travelled and the final displacement.[3]
Show complete worked solution
(a)
$0$-$4\,\mathrm{s}$: displacement increases linearly, so the runner moves forward at constant velocity.
$4$-$7\,\mathrm{s}$: displacement is constant, so the runner is stationary.
$7$-$12\,\mathrm{s}$: displacement decreases linearly and crosses zero, so the runner moves in the negative direction at constant velocity.
$12$-$14\,\mathrm{s}$: displacement is constant again, so the runner is stationary.
Final answer: Moves forward uniformly, rests, returns uniformly past the origin, then rests.
Final answer: $+3.0\,\mathrm{m\,s^{-1}}$ and $-4.0\,\mathrm{m\,s^{-1}}$.
(c)
The runner first travels $12\,\mathrm{m}$.
The return segment is from $+12\,\mathrm{m}$ to $-8\,\mathrm{m}$, a distance of $20\,\mathrm{m}$.
$$d=12+20=\boxed{32\,\mathrm{m}}$$
$$\Delta x=\boxed{-8\,\mathrm{m}}$$
Final answer: $32\,\mathrm{m}$ and $-8\,\mathrm{m}$.
QUESTION 131[10 marks]
Data-basedFoundationCalculatorAbout 12.5 min
A ball is projected from level ground at $22.0\,\mathrm{m\,s^{-1}}$ and $40^\circ$ above the horizontal. Ignore air resistance.
a. Calculate the horizontal and vertical components of the launch velocity.[3]
b. Determine the maximum height.[3]
c. Calculate the flight time and horizontal range.[4]
During a six-second interval, the acceleration of a cart changes as shown. Its velocity at $t=1.0\,\mathrm{s}$ is $2.0\,\mathrm{m\,s^{-1}}$.
a. Estimate the change in velocity from $1.0$ to $6.0\,\mathrm{s}$.[4]
b. Determine the velocity at $t=6.0\,\mathrm{s}$.[2]
Show complete worked solution
(a)
The change in velocity is the area under the acceleration-time graph.
From $1$-$2\,\mathrm{s}$,
$$A_1=\frac{1.0+2.0}{2}(1.0)=1.5\,\mathrm{m\,s^{-1}}$$
From $2$-$4\,\mathrm{s}$,
$$A_2=(2.0)(3.0)=6.0\,\mathrm{m\,s^{-1}}$$
From $4$-$6\,\mathrm{s}$,
$$A_3=\tfrac12(2.0)(3.0)=3.0\,\mathrm{m\,s^{-1}}$$
$$\Delta v=1.5+6.0+3.0=\boxed{10.5\,\mathrm{m\,s^{-1}}}$$
Final answer: $v=-2.0\,\mathrm{m\,s^{-1}}$ and $a=0\,\mathrm{m\,s^{-2}}$.
(c)
At $t=2.0\,\mathrm{s}$, the calculated velocity is $-2.0\,\mathrm{m\,s^{-1}}$, which is not zero.
Zero acceleration means only that the instantaneous rate of change of velocity is zero.
Final answer: Its velocity is $-2.0\,\mathrm{m\,s^{-1}}$; zero acceleration only means the velocity is momentarily not changing.
QUESTION 135[6 marks]
Multi-partStandardCalculatorAbout 7.5 min
A steel ball rolling down a straight ramp passes point P at $1.30\,\mathrm{m\,s^{-1}}$ and point Q at $2.90\,\mathrm{m\,s^{-1}}$. The travel time from P to Q is $1.60\,\mathrm{s}$. Its acceleration is uniform.
a. Calculate the average velocity between P and Q.[2]
b. Determine the distance PQ.[2]
c. Calculate the acceleration.[2]
Show complete worked solution
(a)
For uniform acceleration,
$$\bar v=\frac{u+v}{2}=\frac{1.30+2.90}{2}=\boxed{2.10\,\mathrm{m\,s^{-1}}}$$
A survey boat points perpendicular to a river bank and moves at $4.90\,\mathrm{m\,s^{-1}}$ relative to the water. The current is $2.70\,\mathrm{m\,s^{-1}}$ downstream and the river is $104\,\mathrm{m}$ wide.
a. Calculate the crossing time.[2]
b. Determine the downstream drift.[2]
c. Determine the boat's speed and direction relative to the bank.[4]
Draw a tangent to the curve at $t=10.0\,\mathrm{s}$.
Choose two widely separated points on the tangent and calculate
$$a=\frac{\Delta v}{\Delta t}$$
A correctly drawn tangent gives approximately $-1.2\,\mathrm{m\,s^{-2}}$.
Final answer: Draw a tangent at $10.0\,\mathrm{s}$ and calculate its gradient.
(c)
The cart is still moving in the positive direction, but its velocity is decreasing. The acceleration therefore points in the negative direction.
Final answer: The positive velocity is decreasing.
QUESTION 138[9 marks]
Data-basedAppliedCalculatorAbout 11.2 min
$t/\mathrm{s}$
0
2
4
6
8
10
$v/\mathrm{m\,s^{-1}}$
0
1.4
3.0
6.8
11.0
13.2
A motion sensor records the following velocities for a cart. Assume straight-line changes between consecutive readings.
a. Plot a velocity-time graph for the data.[2]
b. Calculate the maximum average acceleration over any two-second interval.[3]
c. Estimate the displacement during the ten-second interval.[4]
Show complete worked solution
(a)
Plot time on the horizontal axis and velocity on the vertical axis.
Mark $(0,0)$, $(2,1.4)$, $(4,3.0)$, $(6,6.8)$, $(8,11.0)$ and $(10,13.2)$, then join adjacent points with straight lines.
Final answer: A graph through all six data points with straight-line joins.
(b)
The velocity changes over successive $2\,\mathrm{s}$ intervals are $1.4$, $1.6$, $3.8$, $4.2$ and $2.2\,\mathrm{m\,s^{-1}}$.
The largest change is $4.2\,\mathrm{m\,s^{-1}}$.
$$a_{max}=\frac{4.2}{2.0}=\boxed{2.1\,\mathrm{m\,s^{-2}}}$$
Final answer: $2.1\,\mathrm{m\,s^{-2}}$ between $6$ and $8\,\mathrm{s}$.
(c)
Use trapezoids of width $2.0\,\mathrm{s}$.
$$s=\sum\frac{v_i+v_{i+1}}{2}(2.0)$$
$$s=(1.4+4.4+9.8+17.8+24.2)\,\mathrm{m}$$
$$\boxed{s=57.6\,\mathrm{m}}$$
Final answer: $57.6\,\mathrm{m}$
QUESTION 139[8 marks]
Data-basedAppliedCalculatorAbout 10 min
The velocity-time graph represents a delivery vehicle moving along a straight road.
a. Calculate the acceleration during the first $5.0\,\mathrm{s}$.[2]
b. Determine the total distance travelled during the $16\,\mathrm{s}$ shown.[4]
c. Determine the average velocity over the complete interval.[2]
Area from $0$-$5\,\mathrm{s}$:
$$A_1=\tfrac12(5)(18)=45\,\mathrm{m}$$
Area from $5$-$11\,\mathrm{s}$:
$$A_2=(6)(18)=108\,\mathrm{m}$$
Area from $11$-$16\,\mathrm{s}$:
$$A_3=\tfrac12(5)(18)=45\,\mathrm{m}$$
$$d=45+108+45=\boxed{198\,\mathrm{m}}$$
Final answer: $198\,\mathrm{m}$
(c)
The velocity remains non-negative, so displacement equals the $198\,\mathrm{m}$ area.
$$\bar v=\frac{198}{16}=\boxed{12.4\,\mathrm{m\,s^{-1}}}$$
Final answer: $12.4\,\mathrm{m\,s^{-1}}$
QUESTION 140[9 marks]
Multi-partAppliedCalculatorAbout 11.2 min
A car starts from rest, accelerates at $2.40\,\mathrm{m\,s^{-2}}$ for $5.0\,\mathrm{s}$, moves at constant velocity for $8.0\,\mathrm{s}$, and then decelerates uniformly to rest in $3.0\,\mathrm{s}$.
a. Draw the acceleration-time graph.[3]
b. Determine the maximum velocity.[2]
c. Calculate the total distance travelled.[4]
Show complete worked solution
(a)
After the first interval,
$$v=(2.40)(5.0)=12.0\,\mathrm{m\,s^{-1}}$$
The final acceleration is
$$a=\frac{0-12.0}{3.0}=-4.00\,\mathrm{m\,s^{-2}}$$
Draw horizontal lines at $+2.40$ from $0$-$5\,\mathrm{s}$, $0$ from $5$-$13\,\mathrm{s}$ and $-4.00$ from $13$-$16\,\mathrm{s}$.
Final answer: Horizontal sections at $+2.40$, $0$ and $-4.00\,\mathrm{m\,s^{-2}}$ over the three intervals.
(b)
The change in velocity is the area under the acceleration-time graph.
$$\Delta v=(2.40)(5.0)=\boxed{12.0\,\mathrm{m\,s^{-1}}}$$
A car starts from rest. Its speed rises uniformly to $24.0\,\mathrm{m\,s^{-1}}$ in $5.0\,\mathrm{s}$, remains constant for $4.0\,\mathrm{s}$, and then falls uniformly to $8.0\,\mathrm{m\,s^{-1}}$ at $13.0\,\mathrm{s}$.
a. Calculate the acceleration during the first and last intervals.[3]
b. Use the area under the velocity-time graph to determine the distance travelled.[5]
Final answer: $t=2.34\,\mathrm{s}$, $R=38.3\,\mathrm{m}$
QUESTION 143[8 marks]
Multi-partFoundationNo calculatorAbout 10 min
During a flight, the passenger display shows $936\,\mathrm{km\,h^{-1}}$.
a. State whether the displayed quantity is a speed or a velocity and justify your answer.[2]
b. Explain why the value should be interpreted as an instantaneous reading rather than the average speed for the entire journey.[2]
c. Convert the displayed value to $\mathrm{m\,s^{-1}}$.[2]
d. At this speed, calculate the time needed to travel $150\,\mathrm{m}$.[2]
Show complete worked solution
(a)
The display gives magnitude only.
A velocity must include direction, so the displayed quantity is a speed.
Final answer: Speed, because no direction is displayed.
(b)
The screen updates the aircraft's current motion.
The journey includes intervals with different speeds, so this single displayed value is not distance for the whole trip divided by total time.
Final answer: It describes the aircraft at that moment; the speed changes during take-off, cruise and landing.
A survey boat points perpendicular to a river bank and moves at $5.80\,\mathrm{m\,s^{-1}}$ relative to the water. The current is $3.40\,\mathrm{m\,s^{-1}}$ downstream and the river is $128\,\mathrm{m}$ wide.
a. Calculate the crossing time.[2]
b. Determine the downstream drift.[2]
c. Determine the boat's speed and direction relative to the bank.[4]
Final answer: $6.72\,\mathrm{m\,s^{-1}}$ at $30.4^\circ$ downstream from perpendicular
QUESTION 145[9 marks]
Multi-partStandardCalculatorAbout 11.2 min
A rubber ball is released from rest $2.40\,\mathrm{m}$ above a floor and reaches the floor in $0.700\,\mathrm{s}$. After the bounce it rises to a maximum height of $1.10\,\mathrm{m}$. Use $g=9.81\,\mathrm{m\,s^{-2}}$ when needed.
a. Calculate $2.40/0.700$ and state what the result represents.[2]
b. Distinguish between an instantaneous velocity and an average velocity.[2]
c. State the velocity immediately before impact, taking upward as positive.[2]
d. Estimate the velocity immediately after the ball leaves the floor.[3]
Show complete worked solution
(a)
$$\frac{2.40}{0.700}=3.43\,\mathrm{m\,s^{-1}}$$
The displacement divided by the elapsed time is the magnitude of the average velocity for the complete fall.
Final answer: $3.43\,\mathrm{m\,s^{-1}}$, the magnitude of the average velocity during the fall.
(b)
Instantaneous velocity is the velocity at a specified instant.
Average velocity over an interval is
$$\bar v=\frac{\text{total displacement}}{\text{total time}}$$
Final answer: Instantaneous velocity applies at one moment; average velocity is total displacement divided by total time.
(c)
$$v=u+at=0+(-9.81)(0.700)$$
$$\boxed{v=-6.87\,\mathrm{m\,s^{-1}}}$$
The negative sign means downward.
Final answer: $-6.87\,\mathrm{m\,s^{-1}}$
(d)
At the rebound height, $v=0$.
$$v^2=u^2+2as$$
$$0=u^2+2(-9.81)(1.10)$$
$$u=\sqrt{2(9.81)(1.10)}=\boxed{+4.65\,\mathrm{m\,s^{-1}}}$$
Final answer: $+4.65\,\mathrm{m\,s^{-1}}$
QUESTION 146[8 marks]
Multi-partStandardCalculatorAbout 10 min
A survey boat points perpendicular to a river bank and moves at $4.45\,\mathrm{m\,s^{-1}}$ relative to the water. The current is $2.35\,\mathrm{m\,s^{-1}}$ downstream and the river is $92\,\mathrm{m}$ wide.
a. Calculate the crossing time.[2]
b. Determine the downstream drift.[2]
c. Determine the boat's speed and direction relative to the bank.[4]
Final answer: $22.3\,\mathrm{m\,s^{-1}}$ at $55.0^\circ$ below horizontal
QUESTION 150[1 mark]
Multiple choiceAdvancedCalculatorAbout 3 min
The velocity-time graph describes an automated shuttle during a $16.0\,\mathrm{s}$ test.
Determine the total distance travelled and the displacement.
Show complete worked solution
Correct answer: A
Positive area:
$$A_+=\tfrac12(4)(12)+(5)(12)+\tfrac12(3)(12)=24+60+18=102\,\mathrm{m}$$
Negative-area magnitude:
$$A_-=\tfrac12(1)(4)+\tfrac12(3)(4)=2+6=8\,\mathrm{m}$$
$$\text{distance}=A_++A_-=102+8=\boxed{110\,\mathrm{m}}$$
$$\text{displacement}=A_+-A_-=102-8=\boxed{94\,\mathrm{m}}$$
Therefore the correct option is A.
QUESTION 151[1 mark]
Multiple choiceAdvancedCalculatorAbout 3 min
The displacement of a laboratory cart is modelled by $x(t)=0.50t^3-3.0t^2+4.0t$, where $x$ is in metres and $t$ is in seconds.
Determine expressions for the velocity and acceleration.
Show complete worked solution
Correct answer: B
$$v(t)=\frac{dx}{dt}=1.50t^2-6.0t+4.0$$
$$a(t)=\frac{dv}{dt}=3.0t-6.0$$
Therefore the correct option is B.
QUESTION 152[1 mark]
Multiple choiceAdvancedCalculatorAbout 3 min
A ball is projected from level ground at $20.0\,\mathrm{m\,s^{-1}}$ and $35^\circ$ above the horizontal. Ignore air resistance.
Calculate the flight time and horizontal range.
Show complete worked solution
Correct answer: C
$$t=\frac{2u_y}{g}=\frac{2(11.5)}{9.81}=2.34\,\mathrm{s}$$
$$R=u_xt=(16.4)(2.34)=\boxed{38.3\,\mathrm{m}}$$
Therefore the correct option is C.
QUESTION 153[1 mark]
Multiple choiceAdvancedCalculatorAbout 3 min
A rescue package leaves a horizontal platform at $12.8\,\mathrm{m\,s^{-1}}$ from a height of $17.0\,\mathrm{m}$.
Determine the impact speed and direction.
Show complete worked solution
Correct answer: D
$$v_y=gt=9.81(1.86)=18.3\,\mathrm{m\,s^{-1}}$$
$$v=\sqrt{12.8^2+18.3^2}=22.3\,\mathrm{m\,s^{-1}}$$
$$\theta=\tan^{-1}\left(\frac{18.3}{12.8}\right)=\boxed{55.0^\circ}$$
Therefore the correct option is D.
QUESTION 154[1 mark]
Multiple choiceAdvancedCalculatorAbout 3 min
A survey boat points perpendicular to a river bank and moves at $5.35\,\mathrm{m\,s^{-1}}$ relative to the water. The current is $3.05\,\mathrm{m\,s^{-1}}$ downstream and the river is $116\,\mathrm{m}$ wide.
Determine the boat's speed and direction relative to the bank.
Show complete worked solution
Correct answer: A
$$v=\sqrt{5.35^2+3.05^2}=6.16\,\mathrm{m\,s^{-1}}$$
$$\theta=\tan^{-1}\left(\frac{3.05}{5.35}\right)=\boxed{29.7^\circ}$$
Therefore the correct option is A.
QUESTION 155[1 mark]
Multiple choiceAdvancedCalculatorAbout 3 min
A ball is projected from level ground at $28.0\,\mathrm{m\,s^{-1}}$ and $55^\circ$ above the horizontal. Ignore air resistance.
Calculate the flight time and horizontal range.
Show complete worked solution
Correct answer: B
$$t=\frac{2u_y}{g}=\frac{2(22.9)}{9.81}=4.68\,\mathrm{s}$$
$$R=u_xt=(16.1)(4.68)=\boxed{75.1\,\mathrm{m}}$$
Therefore the correct option is B.
QUESTION 156[1 mark]
Multiple choiceAdvancedCalculatorAbout 3 min
A car is $120\,\mathrm{m}$ east of an observer and approaches at a constant velocity of $24.0\,\mathrm{m\,s^{-1}}$ west. It passes the observer, continues for $2.50\,\mathrm{s}$, and then brakes uniformly to rest $132\,\mathrm{m}$ west of the observer.
Choose east as positive and sketch the velocity-time graph from the initial instant until the car stops.
Show complete worked solution
Correct answer: C
The car reaches the observer after
$$t=\frac{120}{24.0}=5.00\,\mathrm{s}$$
Braking begins at $5.00+2.50=7.50\,\mathrm{s}$.
The graph remains at $-24.0\,\mathrm{m\,s^{-1}}$ until $7.50\,\mathrm{s}$ and then rises linearly to zero.
Therefore the correct option is C.
QUESTION 157[1 mark]
Multiple choiceAdvancedCalculatorAbout 3 min
A sample container is released from rest $27.0\,\mathrm{m}$ above the ground. Air resistance is negligible.
Calculate the fall time.
Show complete worked solution
Correct answer: D
$$h=\tfrac12gt^2$$
$$t=\sqrt{\frac{2(27.0)}{9.81}}=\boxed{2.35\,\mathrm{s}}$$
Therefore the correct option is D.
QUESTION 158[1 mark]
Multiple choiceAdvancedCalculatorAbout 3 min
A track cart passes a marker at $9.20\,\mathrm{m\,s^{-1}}$ and then accelerates uniformly at $1.96\,\mathrm{m\,s^{-2}}$ for $9.0\,\mathrm{s}$.
Calculate the displacement during the interval.
Show complete worked solution
Correct answer: A
$$s=ut+\tfrac12at^2$$
$$s=(9.20)(9.0)+\tfrac12(1.96)(9.0)^2=\boxed{162\,\mathrm{m}}$$
Therefore the correct option is A.
QUESTION 159[1 mark]
Multiple choiceAppliedCalculatorAbout 2.5 min
A research satellite follows a circular orbit of radius $7.20\times10^6\,\mathrm{m}$ and completes one orbit every $96.0\,\mathrm{min}$.
Determine its displacement from its starting point after (i) $384\,\mathrm{min}$ and (ii) $432\,\mathrm{min}$.
Show complete worked solution
Correct answer: B
$$\frac{384}{96}=4.00\text{ orbits}$$
The satellite returns to its starting point, so
$$\boxed{\Delta r=0}$$
$$\frac{432}{96}=4.50\text{ orbits}$$
It is diametrically opposite its starting point.
$$|\Delta r|=2r=2(7.20\times10^6)=\boxed{1.44\times10^7\,\mathrm{m}}$$
Therefore the correct option is B.
QUESTION 160[1 mark]
Multiple choiceAppliedCalculatorAbout 2.5 min
The velocity-time graph represents a delivery vehicle moving along a straight road.
Determine the total distance travelled during the $16\,\mathrm{s}$ shown.
Show complete worked solution
Correct answer: C
Area from $0$-$5\,\mathrm{s}$:
$$A_1=\tfrac12(5)(18)=45\,\mathrm{m}$$
Area from $5$-$11\,\mathrm{s}$:
$$A_2=(6)(18)=108\,\mathrm{m}$$
Area from $11$-$16\,\mathrm{s}$:
$$A_3=\tfrac12(5)(18)=45\,\mathrm{m}$$
$$d=45+108+45=\boxed{198\,\mathrm{m}}$$
Therefore the correct option is C.
QUESTION 161[1 mark]
Multiple choiceAppliedCalculatorAbout 2.5 min
$t/\mathrm{s}$
0
2
4
6
8
10
$v/\mathrm{m\,s^{-1}}$
0
1.4
3.0
6.8
11.0
13.2
A motion sensor records the following velocities for a cart. Assume straight-line changes between consecutive readings.
Estimate the displacement during the ten-second interval.
Show complete worked solution
Correct answer: D
Use trapezoids of width $2.0\,\mathrm{s}$.
$$s=\sum\frac{v_i+v_{i+1}}{2}(2.0)$$
$$s=(1.4+4.4+9.8+17.8+24.2)\,\mathrm{m}$$
$$\boxed{s=57.6\,\mathrm{m}}$$
Therefore the correct option is D.
QUESTION 162[1 mark]
Multiple choiceAppliedCalculatorAbout 2.5 min
$t/\mathrm{s}$
0
2
4
6
8
10
12
14
16
$v/\mathrm{m\,s^{-1}}$
0
1.2
2.8
6.5
11.9
17.6
20.1
21.5
22.0
A motion logger records the velocity of a car every $2.0\,\mathrm{s}$. Assume straight-line changes between readings.
Estimate the distance travelled in $16.0\,\mathrm{s}$.
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Correct answer: A
Apply the trapezoidal rule with interval width $2.0\,\mathrm{s}$.
$$s=\sum \frac{v_i+v_{i+1}}{2}(2.0)$$
$$s=1.2+4.0+9.3+18.4+29.5+37.7+41.6+43.5$$
$$s=\boxed{185.2\,\mathrm{m}\approx185\,\mathrm{m}}$$
Therefore the correct option is A.
QUESTION 163[1 mark]
Multiple choiceAppliedCalculatorAbout 2.5 min
An inspection robot has the velocity-time graph shown. Positive velocity is east.
Calculate the displacement during the complete motion.
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Correct answer: B
Eastward area:
$$A_E=\tfrac12(3)(8)+(3)(8)+\tfrac12(3)(8)=48\,\mathrm{m}$$
Westward area magnitude:
$$A_W=\tfrac12(5)(6)+\tfrac12(6)(6)=33\,\mathrm{m}$$
$$\Delta x=48-33=\boxed{15.0\,\mathrm{m}\text{ east}}$$
Therefore the correct option is B.
QUESTION 164[1 mark]
Multiple choiceAppliedCalculatorAbout 2.5 min
A car starts from rest, accelerates at $2.40\,\mathrm{m\,s^{-2}}$ for $5.0\,\mathrm{s}$, moves at constant velocity for $8.0\,\mathrm{s}$, and then decelerates uniformly to rest in $3.0\,\mathrm{s}$.
Calculate the total distance travelled.
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Correct answer: C
$$s_1=\tfrac12(5.0)(12.0)=30.0\,\mathrm{m}$$
$$s_2=(8.0)(12.0)=96.0\,\mathrm{m}$$
$$s_3=\tfrac12(3.0)(12.0)=18.0\,\mathrm{m}$$
$$s=30.0+96.0+18.0=\boxed{144\,\mathrm{m}}$$
Therefore the correct option is C.
QUESTION 165[1 mark]
Multiple choiceAppliedCalculatorAbout 2.5 min
During a six-second interval, the acceleration of a cart changes as shown. Its velocity at $t=1.0\,\mathrm{s}$ is $2.0\,\mathrm{m\,s^{-1}}$.
Estimate the change in velocity from $1.0$ to $6.0\,\mathrm{s}$.
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Correct answer: D
The change in velocity is the area under the acceleration-time graph.
From $1$-$2\,\mathrm{s}$,
$$A_1=\frac{1.0+2.0}{2}(1.0)=1.5\,\mathrm{m\,s^{-1}}$$
From $2$-$4\,\mathrm{s}$,
$$A_2=(2.0)(3.0)=6.0\,\mathrm{m\,s^{-1}}$$
From $4$-$6\,\mathrm{s}$,
$$A_3=\tfrac12(2.0)(3.0)=3.0\,\mathrm{m\,s^{-1}}$$
$$\Delta v=1.5+6.0+3.0=\boxed{10.5\,\mathrm{m\,s^{-1}}}$$
Therefore the correct option is D.
QUESTION 166[1 mark]
Multiple choiceAppliedCalculatorAbout 2.5 min
A rescue package leaves a horizontal platform at $11.0\,\mathrm{m\,s^{-1}}$ from a height of $14.0\,\mathrm{m}$.
Determine the impact speed and direction.
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Correct answer: A
$$v_y=gt=9.81(1.69)=16.6\,\mathrm{m\,s^{-1}}$$
$$v=\sqrt{11.0^2+16.6^2}=19.9\,\mathrm{m\,s^{-1}}$$
$$\theta=\tan^{-1}\left(\frac{16.6}{11.0}\right)=\boxed{56.4^\circ}$$
Therefore the correct option is A.
QUESTION 167[1 mark]
Multiple choiceAppliedCalculatorAbout 2.5 min
A survey boat points perpendicular to a river bank and moves at $4.90\,\mathrm{m\,s^{-1}}$ relative to the water. The current is $2.70\,\mathrm{m\,s^{-1}}$ downstream and the river is $104\,\mathrm{m}$ wide.
Determine the boat's speed and direction relative to the bank.
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Correct answer: B
$$v=\sqrt{4.90^2+2.70^2}=5.59\,\mathrm{m\,s^{-1}}$$
$$\theta=\tan^{-1}\left(\frac{2.70}{4.90}\right)=\boxed{28.9^\circ}$$
Therefore the correct option is B.
QUESTION 168[1 mark]
Multiple choiceAppliedCalculatorAbout 2.5 min
A ball is projected from level ground at $26.0\,\mathrm{m\,s^{-1}}$ and $50^\circ$ above the horizontal. Ignore air resistance.
Calculate the flight time and horizontal range.
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Correct answer: C
$$t=\frac{2u_y}{g}=\frac{2(19.9)}{9.81}=4.06\,\mathrm{s}$$
$$R=u_xt=(16.7)(4.06)=\boxed{67.9\,\mathrm{m}}$$
Therefore the correct option is C.
QUESTION 169[1 mark]
Multiple choiceAppliedCalculatorAbout 2.5 min
A rescue package leaves a horizontal platform at $18.2\,\mathrm{m\,s^{-1}}$ from a height of $26.0\,\mathrm{m}$.
Determine the impact speed and direction.
Show complete worked solution
Correct answer: D
$$v_y=gt=9.81(2.30)=22.6\,\mathrm{m\,s^{-1}}$$
$$v=\sqrt{18.2^2+22.6^2}=29.0\,\mathrm{m\,s^{-1}}$$
$$\theta=\tan^{-1}\left(\frac{22.6}{18.2}\right)=\boxed{51.1^\circ}$$
Therefore the correct option is D.