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.
Answers
Answered by
5
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...!!!..
Answered by
11
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
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