Cambridge International AS Level Physics
Answers to self-assessment questions
Answers to SAQs
Chapter 1
11
10 000 = 6.34 m s–1
1average speed = 1577.53
2
a
b
c
d
e
mm s–1
mph
km s–1
m s–1
km h–1
s
C
–1
m
4average speed = 0.050
0.40 s =–10.125 m s
≈ 0.13 m s
a Constant speed
b Increasing speed (accelerating)
0
a
b
c
d
Displacement
Speed
Velocity
Distance
8Distance d = v × t = 1500 × 0.2 = 300 m.
(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.)
9Time 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.
10
D
t
0
12
a 85 m s–1
b Graph is a straight line through the origin,
with gradient = 85 m s–1.
13
a 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
–1
= 84
4 = 21 km h
14
a Total distance travelled = 3.0 + 4.0 = 7.0 km
b, c The 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(4.0
3.0) = 53° E of N (or 37° N of E)
15
a, b 8.5 km; 48° W of S
6For example, attach a card to a weight and
drop it through a light gate. Alternatively,
attach ticker-tape to the falling mass.
7
B
A
3distance = 12 cm = 120 mm, so
–1
average speed = 120
60 = 2.0 mm s
5
OA: constant speed.
AB: stationary.
BC: reduced constant speed.
CD: running back to gate.
A
48° 45°
8.5 km
W
8.0 km
12.0 km
SE
Sloping sections: bus moving. Horizontal
sections: bus stationary (e.g. at bus stops).
Cambridge International AS and A Level Physics © Cambridge University Press 2014
Cambridge International AS Level Physics
16
17
Answers to self-assessment questions
Swimmer aims directly across river; river
flows at right angles to where she aims. So
resultant velocity is given by geometry:
magnitude2 = 2.02 + 0.82 = 4.64 so
magnitude = 4.64 = 2.154 ≈ 2.2 m s–1
direction = tan–1(0.8
2 ) ≈ 22° to the direct route
(68° to the river bank)
a
44°
vertical
18 ms–1
resultant
25 ms–1
horizontal
17 ms–1
b 17.3 m s–1 ≈ 17 m s–1
c 43.9° ≈ 44° to the vertical
Cambridge International AS and A Level Physics © Cambridge University Press 2014