Question

A car running at the rate of 36km/h is brought to rest in 1 minute 25 seconds. find the retardation prodused and the distance travelled before it come to a halt​

Answer 1

Velocity of the car = 36 km/h = 10 m/s

Time in which car is to be stopped = 2 s

Final velocity of the stopped car = 0 m/s

Using the relation;

v = u + a t

and substituting v = 0 m/s; u = 10 m/s; and t = 2 s,

we get

0 = 10 + 2 a

=> a = - 10/2 = - 5 m/s².

Distance travelled by the car can be obtained using the concept of average velocity.

Averave velocity during the two second braking interval = ½ × (initial velocity at the beginning of the interval + final velocity at the end of the two second interval) = ⅓(10 m/s + 0 m/s) = 5 m/s

Distance travelled at this average speed in 2 second = 5 m/s ×2 = 10 m.