Question

a car is moving with a speed of 36 km per hour comes to rest in 5 second by applying brakes calculate its acceleration and distance covered by the car before coming to rest

Answer 1

u=36 km/hr
=5/18*36
=10 m/s
v=0
t=5 s
a=(v-u)/t
=(0-10)/5
=-10/5
=-2 m/s^2
S=ut+1/2at^2
=10*5+1/2*-2*5*5
=50-25
=25 m
Therefore retardation is 2 m/s^2 and the distance covered by the car before coming to rest is 25 m.

Answer 2

Initial velocity of car=36km/h or 10m/s
Final velocity of car=0m/s
Time taken to come to rest=5s
Acceleration=?
v=u+at
0=10+5a
-10=5a
-2m/s^2=a
Distance travelled by the car:
s=ut+1/2at^2
s=10×5+1/2×-2×25
s=50+(-25)
s=25m
So your answer is- Acceleration of the car=-2m/s. Distance travellby the car=25m.
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