Question

A body with velocity 5 m/s retards at 2m/s the distance travelled between t = 2s to t=3s is?​

Answer 1

here, u= 5m/s

a = -2m/s^2

a= v-u/t

when t = 2 sec

-2 = v1 - 5/2

-4 = v1 - 5

v1 = -1 m/s

when t = 3 sec

-2 = v2-5 / 3

-6 = v2-3

v2 = -3

as s = v/t

therefore, distance travelled between 2 sec and 3 sec

= v2 - v1/t2-t1

s = -3 + 1 /3-2

s = -2/1

s=-2

thus displacement will be 2 km in the backward motion or in the motion away from initial point