Question

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

Answer 1

let the time taken to

travel distance 's' be = t₁

travel distance '2s' be = t₂

travel distance '5s' be = t₃

now,

for a body to travel a distance 's' under uniform acceleration 'a'...

here,

v = u +at

or [as u = 0]

v = 0 + at₁

thus, a = v/t₁

now,

we also have

s = ut + (1/2)at²

or

s = 0 + (1/2)at₁² = (1/2).(v/t₁).t₁²

thus,

s = (1/2)v.t₁

or

time taken in first case

t₁ = 2s / v .......................(1)

..

now,

for the body to travel uniformly a distance of '2s'

the time taken will be

t₂ = 2s/v ..........................(2)

and

..

for the body to travel a distance '5s' under uniform acceleration 'a'...

here,

v = u + at

or [as u = v and v = 0]

0 = v + (-a).t₃

thus, a = v/t₃

now,

we also have

s = ut + (1/2)at²

or

5s = vt₃ + (1/2)at₃² = vt₃ + (1/2).(v/t₃).t₃²

thus,

5s = (1/2)v.t₃

or

time taken in third case

t₃ = 10s / v .......................(3)

.

.

now, average speed = total distance travelled / total time taken

v_av = (s + 2s + 5s) / (t₁ + t₂ + t₃)

= 8s / [(2s/v) + (2s/v) + (10s/3v)] = 8s / (14s/3v)

thus,

v_av = 4v/7 ..........................(4)

and

the maximum speed will be

v_max = v .............................(5)

thus, the ratio will be

v_max / v_av = v / (4v/7)

thus,

v_max / v_av = 7/4