Question

A body starts from rest and travels a distance d with uniform acceleration, then moves uniformly a distance 2d and finally comes to rest after moving further 5d under uniform retardation. the ratio of the average velocity to maximum velocity is:

Answer 1

AB = L, BC = 2L, CD = 5L

it starts from rest and accelerates and then remains in uniform velocity and then retards

this means the maximum velocity will be at point B and let it be v

The velocity time graph is shown below

the area under velocity time graph is displacement and hence we have 3 equation corrosponding to regions AB, BC and CD

v$t_{1}/2$ = L , \quad vt_2 = 2L , \quad vt_3/2 = 5L

t_1+t_2+t_3 = 2L/v +2L/v +6L/v = 10L/v

and the average velocity = tot.displacement/tot. time

\bar{v} = 6L/(t_1+t_2+t_3) = 3v/5

and hence

\bar{v}/v = 3/5