A car is moving with a velocity of 45km/hr. The driver applies brake and the car comes to rest in 3 second. What is the retardation and how far does it move before coming to rest?
Solution:
Initial velocity of car (u) = 45 km/hr = 45 * 1000/3600 = 12.5 m/s
Time taken (t) = 3 s
Since the car stops finally, final velocity of car (v) = 0 m/s
Now,
Retardation (-a) = (u - v)/t
= (12.5 - 0)/3
= 4.17 m/s2
Hence, the magnitude of retardation of the car is 4.17 m/s2.
Let the car move a distance 's' before coming into rest than from equations of motion, we have,
s = ut - 1/2 (a)t2 [acceleration is negative due to retardation]
= (12.5)3 - 1/2 (4.17) (3)2
= 18.375 m
Hence, the car travelled...