Question

A car running 72 km/h is slowed down to 18 km/h by the application of brakes over a distance of 12.5 m. Then total time in which car stops in sec is

Answer 1

As the brakes are applied with the same force so the acceleration will be constant.
Using the equation of motion,
v2=u2+2as⇒102=202+2×a×25⇒50a=100−400⇒a=−30050=−6 m/s2The negative sign shows the retardation.
(i) Let us find out the total time in which the car comes to rest. As the car comes to rest so the final velocity, v = 0
v=u+at⇒0=20−6t⇒t=206=103 s
(ii)
The distance travelled by it,
v2=u2+2as⇒0=202+2×(−6)×s⇒12s=400⇒s=4006=66.67 m