CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK
Exam-style questions and sample answers have been written by the authors. In examinations, the way marks are awarded
may be different.
Coursebook answers
Chapter 1
Self-assessment questions
1
average speed =
2
a
mm s
3
10000
1625.17
= 6.15 m s−1
10 S
loping sections: bus moving; horizontal
sections: bus stationary (e.g., at bus stops)
−1
b
km h-1
c
km s−1
d
m s−1
11 O
A: constant speed; AB: stationary; BC:
reduced constant speed; CD: running back
to gate
s
C
distance = 12 cm = 120 mm
A
so, average speed = 120
= 2.0 mm s−1`
60
0.05 m
0.40 s
4
average speed =
0.013 m s−1
5
a
Constant speed
b
Increasing speed (accelerating)
6
7
8
= 0.0125 m s−1 ≈
0
or example, attach a card to a weight and
F
drop it through a light gate. Alternatively,
attach ticker-tape to the falling mass.
a
Displacement
b
Speed
c
Velocity
d
Distance
distance s = v × t = 1500 × 0.2 = 300 m
t ime taken for orbit is one year = 1 × 365.25 ×
24 × 60 × 60 = 31 557 600 s.
distance travelled = circumference of orbit =
2 × π × 1.5 × 1011 = 9.425 × 1011 m
so, the Earth’s speed = 29.9 km s−1 ≈ 30 km s−1
As the Earth orbits the Sun, its direction of
motion keeps changing. Hence, its velocity
keeps changing. In the course of one year, its
displacement is zero, so its average velocity is
zero.
1
D
t
0
12 a
85 m s
−1
b
13 a
(Remember: the 0.4 s total time is that taken
for the sound waves to travel out and be
reflected back from the surface of the water.)
9
B
Graph
is a straight line through the
origin, with gradient = 85 m s−1
Graph is a straight line for the first 3 h;
then less steep for the last hour
b
car’s speed in first three hours = 23 km h−1
c
car’s average speed in first four hours =
84
4
= 21 km h−1
14 atotal distance travelled = 3.0 + 4.0 =
7.0 km
b, c T
he two parts of the journey are at 90° to
each other, so the resultant displacement
is given by Pythagoras’ theorem.
displacement2 = 3.02 + 4.02 = 25.0, so
displacement = 5.0 km
angle = tan−1
of E
( 34..00 ) = 53° E of N or 37° N
Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside
© Cambridge University Press 2020
CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK
15 a, b8.5 km; 48° W of S or a bearing of 228°
17 a
44°
A
48° 45°
8.5 km
8.0 km
W
magnitude2 = 2.02 + 0.82 = 4.64 so magnitude =
4.64 = 2.154 ≈ 2.2 m s−1
( )
horizontal
17 ms–1
SE
12.0 km
16 S
wimmer aims directly across river; river flows
at right angles to where she aims. So, resultant
velocity is given by geometry:
direction = tan−1 02.8 ≈ 22° to the direct route
(68° to the river bank)
2
resultant
25 ms–1
vertical
18 ms–1
b
17.3 m s−1 ≈ 17 m s−1
c
43.9° ≈ 44° to the vertical
18 a
10 m s−1 North
b
0 m s−1
c
7.1 m s−1 045° or N45°E
d
7.1 m s−1 315° or N45°W
Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside
© Cambridge University Press 2020