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Rectilinear Motion

Question

A car is moving towards check post with velocity of 54 kmh-1. When the car is at a distance of 400 m from the check post, the driver applies brakes which causes a deceleration of 0.3 ms-2. Find the distance (in meter) of car from the check post for 2 minutes after applying the brakes.

Moderate
Solution

Here, u=54×518=15ms1

a=0.3ms2

 v=u+at

or 0=150.3t0

 t0=50s

After 50 s, car permanently comes to rest. 

s=u+v2t=15+02×50=375m

Thus, the distance of car from the check post is

(400375)m=25m



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