Physics Matters for GCE ‘O’ Level (4th Edition): Full Solutions to Textbook Questions Chapter 2
© 2013 Marshall Cavendish International (Singapore) Private Limited
2.1
Chapter 2 Kinematics
Quick Check (page 23)
Speed = distance travelled
time taken
= 200 m
25 s
= 8 m s−1
Velocity = displacement (linear distance)
time taken
= 50 m
25 s
= 2 m s−1
(Note: The displacement of the athlete is the linear distance between her start point and finishing point.)
Quick Check (page 24)
1. Yes, the stone undergoes acceleration when whirled in circles. This is because its direction (and
hence, velocity) is changing constantly. The acceleration is towards the middle of the circular
path.
2. (a) No. The velocity of a space shuttle increases during blasting off — acceleration takes place.
(b) Yes
(c) Yes
(d) No. An MRT train picks up speed when it is leaving a station.
(e) Yes
Quick Check (page 25)
No. The velocity of an object undergoing uniform acceleration increases at a constant rate, while an
object undergoing uniform motion moves with uniform velocity.
Test Yourself 2.1 & 2.2 (page 26)
1. (a) Distance travelled = (4 + 2 + 4 + 2) m = 12 m
(b) Displacement = 0 m from the initial point A
2. Velocity has direction, but speed does not.
3. True. An object moving at a constant velocity has to move in a straight line, otherwise its direction
will change and its velocity will not be a constant.
A4m
2m
4m
2m
N
Physics Matters for GCE ‘O’ Level (4th Edition): Full Solutions to Textbook Questions Chapter 2
© 2013 Marshall Cavendish International (Singapore) Private Limited
2.2
4. (a) Total distance travelled = 50 924 km – 50 780 km = 144 km
Average speed in km h–1 = 144 km
2 h = 72 km h–1
(b)
Average speed in m s–1 = 144 000 m
(2 × 60 × 60) s = 20 m s–1
5. Acceleration is the rate of change of velocity.
6.
a=v−u
t
Quick Check (page 27)
Quick Check (page 30)
The velocity of an object undergoing deceleration decreases with time, while the velocity of an object
experiencing decreasing acceleration still increases with time, but does so at a decreasing rate.
Quick Check (page 31)
No. Since CF is curved and not straight, the car is not moving at a uniform velocity.
D and E lie on a curve with a negative gradient. This means that the car is travelling towards O. The
gradient at D is greater than the gradient at E. The velocity of the car at D is higher than that at E.
From E to F, the gradient is zero. The velocity is neither increasing nor decreasing as the car is at
rest.
Quick Check (page 37)
The upward direction is assigned as the positive direction. Therefore, when the volleyball falls
downwards, its displacement is taken to be negative.

0 Time
Displacement
car in
Figure 2.8
car at
higher
velocity
Velocity
Time
Physics Matters for GCE ‘O’ Level (4th Edition): Full Solutions to Textbook Questions Chapter 2
© 2013 Marshall Cavendish International (Singapore) Private Limited
2.3
Test Yourself 2.3 (page 38)
1. Time
interval
Type of
motion Average velocity
0–10 Non-uniform
velocity Average velocity = total displacement
total time taken = 50 m − 0 m
10 s = 5 m s–1
10–30 Uniform
velocity Average velocity = 260 m − 50 m
20 s = 10.5 m s–1
30–35 Non-uniform
velocity Average velocity = 280 m − 260 m
5 s = 4 m s–1
35–40 At rest
Average velocity = 280 m − 280 m
5 s = 0 m s–1
2. We can tell if an object is stationary by checking if its displacement–time graph is horizontal
(i.e. its displacement is constant with time).
3. The velocity of an object is given by the gradient of its displacement–time graph. For an object
moving at non-uniform velocity, its velocity (i.e. instantaneous velocity) is given by the gradient of
the tangent of the displacement–time graph at that particular instant of time.
4. Displacement–time graph (Figure 2.20)
(a) From O to A, the car moves with increasing velocity (i.e. acceleration).
(b) From A to B, the car moves with decreasing velocity (i.e. deceleration).
(c) From B to C, the car is at rest.
(d) From C to D, the car moves with constant velocity.
Velocity–time graph (Figure 2.21)
(a) From O to A, the car moves with increasing acceleration.
(b) From A to B, the car moves with decreasing acceleration.
(c) From B to C, the car moves with constant velocity.
(d) From C to D, the car moves with constant acceleration.
Test Yourself 2.4 (page 44)
1. (a)
(b) Gradient of v–t graph = acceleration of free fall g
= v − 0 m s–1
5s
Velocity v of object just before it hits the ground = 5 s × g
= 5 s × 10 m s–2
= 50 m s–1
(c) Height = area under v–t graph
= 12
vt
= 12
(50 m s–1)(5 s)
= 125 m
Velocity/m s–1
Time/s
5
Physics Matters for GCE ‘O’ Level (4th Edition): Full Solutions to Textbook Questions Chapter 2
© 2013 Marshall Cavendish International (Singapore) Private Limited
2.4
2. A feather is less dense than a hammer. Therefore, a falling feather experiences significantly
higher air resistance than a falling hammer over the same period of time. The feather thus
reaches terminal velocity faster than the hammer.
IT Learning Room (page 45)
Part 1
4. Position slider
Moving the Position slider from 1.20 m to –1.20 m changes the position of the man. At positive
values, the man stands to the right of the origin (i.e. 0 m). At negative values, the man stands to
the left of the origin. The man remains stationary as time passes.
Velocity slider
Moving the Velocity slider from 1.20 m s–1 to –1.20 m s–1 changes the velocity of the man.
Decreasing the numerical value from 1.20 m s–1 to 0 m s–1 reduces the speed of the man.
Increasing the numerical value increases his speed. The positive or negative sign denotes the
direction in which the man travels — a positive sign indicates that his movement is to the right of
the origin and a negative sign indicates that his movement is to the left.
Acceleration slider
Moving the Acceleration slider from 1.20 m s–2 to –1.20 m s–2 changes the acceleration of the
man. Decreasing the numerical value from 1.20 m s–2 to 0 m s–2 reduces the acceleration of the
man. Increasing the numerical value increases the acceleration of the man. The positive or
negative sign denotes the direction in which the man travels — a positive sign indicates that his
movement is to the right of the origin and a negative sign indicates that his movement is to the
left.
Physics Matters for GCE ‘O’ Level (4th Edition): Full Solutions to Textbook Questions Chapter 2
© 2013 Marshall Cavendish International (Singapore) Private Limited
2.5
Part 2
4. Graph generated based on scenario 1
Simulation images © PhET Interactive Simulations, University of Colorado
(http://phet.colorado.edu)
Physics Matters for GCE ‘O’ Level (4th Edition): Full Solutions to Textbook Questions Chapter 2
© 2013 Marshall Cavendish International (Singapore) Private Limited
2.6
Graph generated based on scenario 2
Simulation images © PhET Interactive Simulations, University of Colorado
(http://phet.colorado.edu)
Part 3
2. Find the gradient of the displacement–time graph and use it to plot the velocity–time graph.
Physics Matters for GCE ‘O’ Level (4th Edition): Full Solutions to Textbook Questions Chapter 2
© 2013 Marshall Cavendish International (Singapore) Private Limited
2.7
Get It Right (page 46)
(a) False
Velocity is a vector, and hence has magnitude and direction.
(b) False
An object travelling at a uniform velocity has a displacement–time graph that has a constant
gradient. The gradient can be negative or positive.
(c) False
When the speed of an object changes at a uniform rate and the object moves in a constant
direction, its acceleration remains constant. When the speed of an object changes at a nonuniform
rate and/or its direction changes, its acceleration changes.
(d) False
The area under a velocity–time graph of an object gives its displacement.
(e) False
The velocity–time graph of a free-falling object has a positive, constant gradient.
(f) True
Let’s Review (pages 47–48)
Section A: Multiple-Choice Questions
1. B
Average speed = total distance travelled
total time taken
Total distance travelled = average speed × total time taken
Total distance travelled = 35 km h–1 × 45
60 h = 26.25 km
2. D
a=v−u
t
= 13 m s–1 − 5 m s–1
4.0 s
= 2.00 m s–2
3. A
a=v−u
t
−10 m s–2 = 0 m s–1 − 1.2 m s–1
t
t = 0.12 s
4. D
A velocity–time graph with a negative gradient shows that an object is decelerating.
A changing gradient shows that the object is undergoing a non-uniform deceleration.
5. C
In vacuum, an object falls with a constant acceleration of 10 m s–2 towards the ground. In the
presence of air resistance, the falling object will accelerate at a slower rate and will eventually
reach a constant velocity known as terminal velocity.