mumtaz jarra (10 )

MOTION
EXAMPPLE 8.7
QUESTION : the brakes applied to a car produxe an acxeration of 6m s in the opposite direction to the motion .if the car takes takes 2 s to stop after the application of brakes ccalculate the distance it travlea during thia time.
ANSWER : a : 6 m s , t : 2 s , v : 0m s
v = u +at
0 = u + (-6 s) * 2 s
or u = 12 m s

s= ut + 1/2at2
= 12 ms * 2ms +1/2 (6ms) (2ms)
=24 m - 12 m
= 12 m
EXAMPLE 8.6
QUESTION : a car accekerates uniformoly from 18 km / h to 36 km / h in 5 sec.cakculate the accekeration nd the distance covered by the var in tgat time
answer. u = 18 km / h
= 5 m / sec
v = 36 km / h
= 10 m / sec
t = 5 sec
(i) a = u-v
---
t
= 10 m s - 5m s
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5s
= 1 m s
(ii) s = ut + 1/2 at2
= 5 m s * 5 s * 1/2 * 1ms * 5 s
= 25 m + 12.5 m
= 37.5 m