Cambridge International AS Level Physics
Answers to end-of-chapter questions
Answers to EOC questions
Chapter 1
1
a distance = speed × time
[1]
× 2 [1]
= 120
60
= 4.0 km
[1]
b The car’s direction of motion keeps changing.
Hence its velocity keeps changing. In the
course of one lap, its displacement is zero, so
its average velocity is zero.
[2]
c distance travelled in 1 minute
= 0.5 × circumference
[1]
but displacement = diameter of track
[1]
circumference
=
[1]
π
m
= 4000
π = 1270 m
2
3
5
6s / m38 40
a distance in car = 0.25 × 60 = 15 km
[1]
total distance = 2.2 + 15 = 17.2 km
[1]
b By Pythagoras’ theorem, displacement
= 2.22 + 152[1]
= 15 200 m
[1]
–1 2.2
at an angle = tan ( 15 )[1]
= 8° E of N
[1]
–1
2200
c time for 2.2 km at 2.0 m s = 2 = 1100 s [1]
total time = 1100 + 900 = 2000 s
[1]
distance
d average speed = time [1]
200
= 17
2000 [1]
= 8.6 m s–1[1]
[1]
e average velocity = displacement
time
15
200
[1]
= 2000 –1
= 7.6 m s [1]
boy
36 35
girl
30
25
20
15
10
5
0
[1]
a By Pythagoras’ theorem,
distance2 = 6002 + 8002 m2[1]
distance = 1 000 000 = 1000 m
[1]
–1 800
b angle at B = tan ( 600 )[1]
displacement = 1000 m at an angle 53° E of N
[1]
c velocity = 1000
60 [1]
–1
= 16.7 m s [1]
at an angle 53° E of N
[1]
a distance in a (particular) direction
[1]
b when athlete returns to his original position
or the start
[1]
(direct) distance from original position zero
[1]
0
1
2
3
4
5
6
7
8
9 10 11 12
t/s
a straight line from t = 0, s = 0 to t = 12, s = 36[1]
b straight line from t = 0, s = 0 to t = 5, s = 10 [1]
straight line from t = 5, s = 10 to t = 12, s = 38
[1]
c 10 s where the graphs cross
[1]
7
a Each second, it travels a constant distance.[1]
At least two examples:
108 – 84 = 24, 84 – 60 = 24, 60 – 36 = 24 cm[1]
24 b s = dt = 0.1
[1]
–1
240 cm s [1]
c 108 + 2 × 24
[1]
156 cm [1]
8
a Vector quantities have direction, and scalar
quantities do not.
[1]
One example of a vector, e.g. velocity,
acceleration, displacement, force
[1]
One example of a vector, e.g. speed, time,
mass, pressure
[1]
100 km h–1
N
resultant
500 km h–1
4resultant velocity = 1.02 + 2.402[1]
= 2.6 m s–1[1]
at an angle of tan–1( 1.0
2.4 )[1]
= 23° E of N
[1]
Cambridge International AS and A Level Physics © Cambridge University Press 2014
Cambridge International AS Level Physics
Answers to end-of-chapter questions
b Correct vectors drawn and labelled
[1]
Scale stated and diagram of sufficient size
[1]
Resultant velocity 510 (±10) km h–1[1]
11° W of N or a bearing of 349° (±3°)
[1]
c 0.25 × 510 = 128 ≈ 130 km 11° W of N
9
a
velocity of aircraft
B
7.5 m s–1
15 m s–1
A
Correct vector diagram
Velocity of aircraft in still air in easterly
direction or calculation
5000
b t = 5000
15 = 333 s or 13.5 = 370 s
[1]
[1]
[1]
total time = 703 or 704 s or 703.7 s
[1]
–1
000
average speed = 10
703.7 = 14.2 m s [1]
Cambridge International AS and A Level Physics © Cambridge University Press 2014