Question

A vehicle moves at a speed of 40km/hr,it stopped by applying brakes which produces a uniform acceleration of -0.6m/s^2.How much distance will the vehicle move before coming to stop.

Answer 1

a=-0.6m/s²
u=40km/hr=40×1000/60×60=100/9m/s

Here brakes are applied therefore final velocity of car is 0.
We know that,
v=u+at

0=100/9-0.6t

0.6t=100/9

t=100/5.4

t=1000/54=500/27 seconds
We also know that,
S=ut+1/2at²

S=(100/9)(500/27)+1/2(-0.6)(500/27)(500/27)

S=50000/243-150000/1458

S=205.77-102.89

S=102.8m

Hence the distance moved by the vehicle before coming to stop is 102.8 metres.

Answer 2

40kmh = 100/9 m/s

2as = v^2-u^2
2 (-0.6)(s) = 0- 100^2/9^2
-1.2s=-10000/81
s=102.8 m