Page 11 - Pra U STPM 2021 Penggal 1 - Physics

Physics Term 1 STPM Chapter 2 Kinematics

Example 4

The graph shows the speeds of two cars A and B Speed / m s –1
which are travelling in the same direction
over a period of 80 s. Car A travelling at a constant 30 B
speed of 20 m s overtakes car B at time t = 0.
–1
2 In order to catch up with car A, car B immediately 20 A
accelerate uniformly for 30 s to reach a constant
speed of 30 m s .
–1
(a) How far does car A travel during the first 30 s?
(b) Calculate the acceleration of car B in the first 30 s. 5
(c) What is the distance travelled by car B in this 0 Time / s
time? 30 80
(d) What additional time will it take for car B to catch up with car A after car A passes car B?
(e) How far would each car have travelled since t = 0?
(f) What is the maximum distance between the cars before car B catches up with car A?

Solution:
(a) Distance travelled by car A during the first 30 s = 30 × 20
= 600 m
(b) Acceleration of car B = Gradient of graph
30 – 5
=
30
= 0.833 m s –2
(c) Distance travelled by car B during the first 30 s
= Area under the graph from t = 0 to t = 30 s
= 1 (5 + 30) × 30
2
= 525 m
(d) After 30 s, car A is ahead of car B by (600 – 525) m = 75 m
For car B to catch up with car A,
Further distance travelled by car B – Further distance travelled by car A = 75 m
30t – 20t = 75
t = 7.5 s
Hence, car B catches up with car A (30 + 7.5) s from when car A passes car B, i.e. 37.5 s.
(e) Distance travelled by each car = 20 × 37.5 m
= 750 m
(f) The two cars are furthest apart when the two graphs intercept at P.
At P, the difference (Area under graph A – Area under graph B) is maximum.
Refering to the figure:
x = 10 x + y = 30 Speed / m s –1
y 15
2 2
x = y y + y = 30 30 B
3 3
3
y = × 30 P
5 20 y A
= 18 s x
15
Hence, maximum separation = Shaded area
1
= × 18 × 15 5
2 Time / s
= 135 m 0 30 80

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02 STPM PHY T1.indd 40 4/9/18 8:19 AM