Page 189 - Physics

Linear motion

Solution Therefore, using

2
2
At the beginning: v - u = 2 as
v 2 u - 2
u = 0 m/s s = 2a
a = 2 m/s 2
0 12 (m/s)- 2 2
t = 6 s = 2 ´ 2.4) m (- /s 2

u
Substituting in v =+ , t a
= 30 m
2
v = 0 m/s + 2 m/s 6 s (a) Total distance covered
= 12 m/s = 36 m + 360 m + 30 m = 426 m

Distance covered initially will be: (b) Maximum speed 12 m/s.=

1
s ut= + at 2 Exercise 8.3
2
1 1. A train travelling at 30 km/h stops
22
s = 0 m/s 6 s´ + ´ 2 m/s ´ 2 6 s
2 when its brakes are applied. The train
2
= 36 m suffers a deceleration of 2 m/s .
Therefore, during the second stage (a) How long does the train take to

u = 12 m/s (constant ) come to rest?
(b) What is its fi nal velocity?
t = 30 s

Therefore, distance covered ut= ´ 2. An object travelling at 10 m/s
accelerates at 4 m/s for 8 seconds.
2
= 12 m/s 30´ s (a) Calculate the fi nal velocity.

= 360 m (b) How far does it travel for
In the third step: 8 seconds?

u = 12 m/s 3. A car moves with a uniform velocity of

12 m/s for 6 seconds. It accelerates at
v = 0 m/s (stopped) 2.0 m/s for 4 seconds. It then travels
2
t = 5 s for 2 more seconds with uniform
velocity. The car fi nally decelerates
Therefore, acceleration is given by: to a stop in 15 seconds. Calculate:

v – u 0 m/s 12 m/s
-
a = = = - 2.4 m/s 2 (a) the distance travelled in
t 5s 5 seconds.

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Physics Form 1 Final.indd 183 16/10/2024 20:58