End of chapter exercises
Give one word/term for the following descriptions.

The shortest path from start to finish.

A physical quantity with magnitude and direction.

The quantity defined as a change in velocity over a time period.

The point from where you take measurements.

The distance covered in a time interval.

The velocity at a specific instant in time.
a) Displacement
b) Vector
c) Acceleration
d) Reference point
e) Velocity
f) Instantaneous velocity
Choose an item from column B that match the description in column A. Write down only the letter next to the question number. You may use an item from column B more than once.

a) Displacement
b) Acceleration
c) velocity
d) velocity
Indicate whether the following statements are TRUE or FALSE. Write only “true” or “false”. If the statement is false, write down the correct statement.

A scalar is the displacement of an object over a time interval.

The position of an object is where it is located.

The sign of the velocity of an object tells us in which direction it is travelling.

The acceleration of an object is the change of its displacement over a period in time.
a) False. An example of a vector is the displacement of an object over a time interval. An example of a scalar is the distance of an object over a time interval.
b) True
c) True
d) False. The acceleration of an object is the change of its velocity over a period of time.
[SC 2003/11] A body accelerates uniformly from rest for ${t}_{0}$ seconds after which it continues with a constant velocity. Which graph is the correct representation of the body's motion?




(a) 
(b) 
(c) 
(d) 
Graph b. Graph d does not have uniform acceleration, graph a and c do not have constant velocity after time ${t}_{0}$ . Graph b has uniform acceleration and then constant velocity.
[SC 2003/11] The velocitytime graphs of two cars are represented by $P$ and $Q$ as shown
The difference in the distance travelled by the two cars (in m) after 4 s is $...$

12

6

2

0
[IEB 2005/11 HG] The graph that follows shows how the speed of an athlete varies with time as he sprints for 100 m.
Which of the following equations can be used to correctly determine the time t for which he accelerates?

$100=\left(10\right)\left(11\right)\frac{1}{2}\left(10\right)t$

$100=\left(10\right)\left(11\right)+\frac{1}{2}\left(10\right)t$

$100=10t+\frac{1}{2}\left(10\right){t}^{2}$

$100=\frac{1}{2}\left(0\right)t+\frac{1}{2}\left(10\right){t}^{2}$
[SC 2002/03 HG1] In which one of the following cases will the distance covered and the magnitude of the displacement be the same?

A girl climbs a spiral staircase.

An athlete completes one lap in a race.

A raindrop falls in still air.

A passenger in a train travels from Cape Town to Johannesburg.
In part b, the athlete ends where he started and so his displacement is 0, while his distance is some number. In part a, the displacement will be less than the distance. In part d, the track is not straight all the way and so the displacement and distance will be different.
Part c however will have the same distance and displacement. The raindrop falls vertically down, it ends in a different place from where it started, and moves in a straight line (straight down). This gives the distance that it falls and the displacement to be the same.
[SC 2003/11] A car, travelling at constant velocity, passes a stationary motor cycle at a traffic light. As the car overtakes the motorcycle, the motorcycle accelerates uniformly from rest for 10 s. The following displacementtime graph represents the motions of both vehicles from the traffic light onwards.

Use the graph to find the magnitude of the constant velocity of the car.

Use the information from the graph to show by means of calculation that the magnitude of the acceleration of the motorcycle, for the first 10 s of its motion is 7,5 m·s^{−2}.

Calculate how long (in seconds) it will take the motorcycle to catch up with the car (point $X$ on the time axis).

How far behind the motorcycle will the car be after 15 seconds?
a) The magnitude of the constant velocity of the car is the same as the gradient of the line:
$m=\frac{{y}_{2}{y}_{1}}{{x}_{2}{x}_{1}}$ 
$m=\frac{3000}{100}$ 
$m=30$ 
So the magnitude of the constant velocity is: $30m\cdot {s}^{1}$
b) We can use the equations of motion. The total distance covered in 10 s is 375 m and the time taken is 10 s. The acceleration is:
$\Delta x={v}_{i}t+\frac{1}{2}a{t}^{2}$ 
$375=0\left(10\right)+\frac{1}{2}a\left({10}^{2}\right)$ 
$375=50a$ 
$a=7,5m\cdot {s}^{2}$ 
c) We use the equations of motion to find the displacement of the motorcycle:
$\Delta x={v}_{i}t+\frac{1}{2}a{t}^{2}$ 
$=0+\frac{1}{2}(7,5{t}^{2})$ 
$=3,75{t}^{2}$ 
And the cars displacement is:
$\Delta x={v}_{i}t+\frac{1}{2}a{t}^{2}$ 
$=30t+0$ 
$=30t$ 
At point X the displacement of the car will be equal to the displacement of the motorcycle:
$3,75{t}^{2}=30t$ 
$3,75{t}^{2}30t=0$ 
$t\left(t8\right)=0$ 
So t = 0 or t = 8. Since t is greater than 0, t = 8 s.
So the motorcycle will take 8 s to catch up with the car.
d) After 15 s the car has travelled a distance of:
$\Delta x={v}_{i}t+\frac{1}{2}a{t}^{2}$ 
$\Delta x=\left(30\right)\left(15\right)+0$ 
$\Delta x=450m$ 
At 10 s the motorcycle will have travelled 375 m (read off graph)
So its velocity will be:
${v}_{f}={v}_{i}+at$ 
${v}_{f}=0+(7,5)\left(10\right)$ 
${v}_{f}=75m\cdot {s}^{1}$ 
And from 10 to 15 s the distance will be:
$\Delta x={v}_{i}t+\frac{1}{2}a{t}^{2}$ 
$\Delta x=\left(75\right)\left(5\right)+0$ 
$\Delta x=375m$ 
The total distance travelled is:
$375+375=750m$
So the motorcycle will have travelled $750450=300m$ further than the car. The car will be 300 m behind the motorcycle after 15 s.
[IEB 2005/11 HG] Which of the following statements is true of a body that accelerates uniformly?

Its rate of change of position with time remains constant.

Its position changes by the same amount in equal time intervals.

Its velocity increases by increasing amounts in equal time intervals.

Its rate of change of velocity with time remains constant.
Its rate of change of velocity with time remains constant.
[IEB 2003/11 HG1] The velocitytime graph for a car moving along a straight horizontal road is shown below.
Which of the following expressions gives the magnitude of the average velocity of the car?

$\frac{\text{Area}\phantom{\rule{4.pt}{0ex}}\text{A}}{t}$

$\frac{\text{Area}\phantom{\rule{4.pt}{0ex}}\text{A}\phantom{\rule{0.277778em}{0ex}}+\phantom{\rule{0.277778em}{0ex}}\text{Area}\phantom{\rule{4.pt}{0ex}}\text{B}}{t}$

$\frac{\text{Area}\phantom{\rule{4.pt}{0ex}}\text{B}}{t}$

$\frac{\text{Area}\phantom{\rule{4.pt}{0ex}}\text{A}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{Area}\phantom{\rule{4.pt}{0ex}}\text{B}}{t}$
[SC 2002/11 SG] A car is driven at 25 m·s^{−1} in a municipal area. When the driver sees a traffic officer at a speed trap, he realises he is travelling too fast. He immediately applies the brakes of the car while still 100 m away from the speed trap.

Calculate the magnitude of the minimum acceleration which the car must have to avoid exceeding the speed limit, if the municipal speed limit is 16,6 m·s^{−1}.

Calculate the time from the instant the driver applied the brakes until he reaches the speed trap. Assume that the car's velocity, when reaching the trap, is 16,6 m·s^{−1}.
a) To avoid exceeding the speed limit his final velocity must be $16,6m\cdot {s}^{1}$ or less. The acceleration is:
${v}_{f}^{2}={v}_{i}^{2}+2a\Delta x$ 
${(16,6)}^{2}={\left(25\right)}^{2}+2\left(100\right)a$ 
$275,56=625+200a$ 
$a=1,75m\cdot {s}^{2}$ 
b)
${v}_{f}={v}_{i}+at$ 
$16,6=25+(1,75)t$ 
$1,75t=8,4$ 
$t=4,8s$ 
A traffic officer is watching his speed trap equipment at the bottom of a valley. He can see cars as they enter the valley 1 km to his left until they leave the valley 1 km to his right. Nelson is recording the times of cars entering and leaving the valley for a school project. Nelson notices a white Toyota enter the valley at 11:01:30 and leave the valley at 11:02:42. Afterwards, Nelson hears that the traffic officer recorded the Toyota doing 140 km·hr^{−1}.

What was the time interval ($\Delta t$) for the Toyota to travel through the valley?

What was the average speed of the Toyota?

Convert this speed to km $\xb7$ hr ${}^{1}$.

Discuss whether the Toyota could have been travelling at 140 km·hr^{−1} at the bottom of the valley.

Discuss the differences between the instantaneous speed (as measured by the speed trap) and average speed (as measured by Nelson).
a) 72 s
b) The Toyota covered a distance of 2 km in 72 s. This is 2 000 m in 72 s. The average speed is:
$\text{speed}=\frac{\text{distance}}{\text{time}}$ 
$s=\frac{2000}{72}$ 
$s=27,78m\cdot {s}^{1}$ 
c) $\frac{27,78\times 3600}{1\times 1000}=100km\cdot {h}^{1}$
d) The Toyota could have been travelling at $140km\cdot {h}^{1}$ at the bottom of the valley. The Toyota may have been accelerating towards the speed trap and then started slowing down. The average speed that Nelson measured does not take changes in acceleration into account.
e) Instantaneous speed is the speed at a specific moment in time. Average speed is the total distance covered divided by the time taken. Average speed assumes that you are travelling at the same speed for the entire distance while instantaneous speed looks at your speed for a specific moment in time.
[IEB 2003/11HG] A velocitytime graph for a ball rolling along a track is shown below. The graph has been divided up into 3 sections, A, B and C for easy reference. (Disregard any effects of friction.)

Use the graph to determine the following:

the speed 5 s after the start

the distance travelled in Section A

the acceleration in Section C


At time ${t}_{1}$ the velocitytime graph intersects the time axis. Use an appropriate equation of motion to calculate the value of time ${t}_{1}$ (in s).

Sketch a displacementtime graph for the motion of the ball for these 12 s. (You do not need to calculate the actual values of the displacement for each time interval, but do pay attention to the general shape of this graph during each time interval.)
ai) 5 s after the start the ball's speed is $0,6m\cdot {s}^{1}$
ii) The distance travelled is the area under the graph. The distance is:
$area=\frac{1}{2}\left(5\right)(0,6)=1,5m$
iii) The acceleration is the slope of the graph.
$m=\frac{{y}_{2}{y}_{1}}{{x}_{2}{x}_{1}}$ 
$m=\frac{0,20,6}{1210}$ 
$m=\frac{0,8}{2}$ 
$m=0,4$ 
The acceleration is:
$0,4m\cdot {s}^{2}$
b) At time t1 the final velocity is 0. The initial velocity is 0,6. The acceleration is 0,4.
${v}_{f}={v}_{i}+at$ 
$0=0,60,4t$ 
$t=1,5s$ 
So t1 is:
${t}_{1}=1,5+10=11,5s$
c)
In towns and cities, the speed limit is 60 km·hr^{−1}. The length of the average car is 3,5 m, and the width of the average car is 2 m. In order to cross the road, you need to be able to walk further than the width of a car, before that car reaches you. To cross safely, you should be able to walk at least 2 m further than the width of the car (4 m in total), before the car reaches you.

If your walking speed is 4 km·hr^{−1}, what is your walking speed in m·s^{−1}?

How long does it take you to walk a distance equal to the width of the average car?

What is the speed in m·s^{−1} of a car travelling at the speed limit in a town?

How many metres does a car travelling at the speed limit travel, in the same time that it takes you to walk a distance equal to the width of car?

Why is the answer to the previous question important?

If you see a car driving toward you, and it is 28 m away (the same as the length of 8 cars), is it safe to walk across the road?

How far away must a car be, before you think it might be safe to cross? How many carlengths is this distance?
a) Your walking speed is: $\frac{4km}{1hour}=\frac{4\times 1000}{1\times 60\times 60}=\frac{4000}{3600}=1,11m\cdot {s}^{1}$
b)
$speed=distanc\frac{e}{t}ime$ 
$1,11=\frac{2}{t}$ 
$t=1,8s$ 
c) The speed of a car travelling at the speed limit is:
$\frac{60km}{1hour}=\frac{60\times 1000}{1\times 60\times 60}=16,67m\cdot {s}^{1}$
d)
$speed=distanc\frac{e}{t}ime$ 
$16,67=\frac{x}{1},8$ 
$x=30m$ 
e) It helps determine the safe distance of a car.
f) No. In the time it takes you to cross the car will have travelled 30 m and have hit you.
g) 60 m to be certain you will make it. This is 18 car lengths. You walk 2 m in 1,8 s. So in order to cover 4 m safely the car must be 60 m away.
A bus on a straight road starts from rest at a bus stop and accelerates at 2 m·s^{−2} until it reaches a speed of 20 m·s^{−1}. Then the bus travels for 20 s at a constant speed until the driver sees the next bus stop in the distance. The driver applies the brakes, stopping the bus in a uniform manner in 5 s.

How long does the bus take to travel from the first bus stop to the second bus stop?

What is the average velocity of the bus during the trip?
We draw a rough sketch of the problem and define a positive direction:
a) We first find the time taken for the bus to reach a speed of $20\text{m}\cdot {\text{s}}^{1}$:
${v}_{f}={v}_{i}+at$ 
$20=0+2t$ 
$t=10\text{s}$ 
Now we have the time for each part of the journey and so we can add all the times together to get the total time:
$10+20+5=35\text{s}$
The bus takes 35 s to get from one bus stop to the next.
b) Average velocity is the displacement divided by the total time. So we need to find the displacement of the bus.
For the first part of the journey the displacement is:
$\Delta x={v}_{i}\left(t\right)+\frac{1}{2}a{t}^{2}$ 
$\Delta x=0\left(10\right)+\frac{1}{2}\left(2\right)\left({10}^{2}\right)$ 
$\Delta x=+100\text{m}$ 
For the second part of the journey the displacement is:
The bus is traveling at a constant speed of $20\text{m}\cdot {\text{s}}^{1}$ for 20 s. The displacement is:
$(20\text{m}\cdot {\text{s}}^{1})\left(20\right)=+400\text{m}$
For the third part of the journey the bus starts with a speed of $20\text{m}\cdot {\text{s}}^{1}$ and ends with a speed of $0\text{m}\cdot {\text{s}}^{1}$ . The displacement is:
$\Delta x=\frac{{v}_{i}+{v}_{f}}{2}\left(t\right)$ 
$\Delta x=\frac{20+0}{2}\left(5\right)$ 
$\Delta x=+50\text{m}$ 
The total displacement is:
$100+400+50=+550\text{m}$
So the average velocity is:
$\frac{550}{35}=15,71\text{m}\cdot {\text{s}}^{1}$