A car moving along a straight highway with speed of 126 km/hr is brought to a stop within a distance of 200 m . What is the retardation of the car (assumed uniform) and how long does it take for the car to stop?

We have,

Initial velocity of the car (u) = 126 km/hr = 35 m/s
Distance travelled by the car before it comes to rest (s) = 200 m
Final velocity of the car (v) = 0 (Since, it’s at rest)
Now, using the 3rd equation of motion,

v2 – u2 = 2as
a = (v2 – u2)/2s
   = (0 – 352)/(2 × 200)
   = –1225/400
   = –3.06 m/s2

The retardation of the car is 3.06 m/s2
Let’s assume the time taken by the car to come to rest as t sec
Using the 1st equation of motion, we have
v = u + at
0 = 35 – 3.06 × t
t = 35/3.06
  = 11.43 sec

Hence, the car will take 11.43 seconds for it to stop.

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