Question

A car is travelling with a speed of 36km/hr.The driver applied brakes and retards the car uniformly. The car stopped in 5 sec .Find the acceleration of car ,distance before it stops after applying the brakes.​

Answer 1

Given,

u=36km/hr

=36×5/18=10m/s

v=0m/s (since the car stopped after 5 sec)

t=5 second

To find, acceleration (a) and displacement (s)

We know, v=u+at

Therefore, 10=0+a×5

=10=5a

or a=10/5=2m/s^2

Also, s=ut+1/2at^2

Therefore, s=0×5+1/2×2×5^2

s=5×5=25m

Implies, Acceleration is 2m/s^2 and Distance is 25 m

Hope it helps u...!!!..

Answer 2

Explanation:

u=36km/hr

=36x5/18=10m/s

v=0m/s (since the car stopped after 5 sec)

t=5 second

To find, acceleration (a) and displacement (s)

We know, v=u+tat

Therefore, 10=0+ax5

=10=5a

or a=10/5=2m/s"2

Also, s=ut+1/2at"2

Therefore, s=0x5+1/2x2x5/2

s=5x5=25m

Implies, Acceleration is 2m/s”*2 and Distance

is 25m