Question

Q. A particle start from rest and transverse a distance L with uniforms acceleration,then moves uniformly over a further distance 2L and finally comes to rest after moving a further distance 3L under uniform retardation. Assuming entire motion to be rectilinear motion the ratio of average speed over journey to the maximum speed on its ways is

Answer 1

let the time taken to

travel distance 'l' be = t1

travel distance '2l' be = t2

travel distance '3l' be = t3

now,

for a body to travel a distance ‘l’ 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)at^2

or

l = 0 + (1/2)at1^2 = (1/2).(v/t1).t1^2

thus,

l = (1/2)v.t1

or

time taken in first case

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

..

now,

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

the time taken will be

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

and

..

for the body to travel a distance '3l' 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)at^2

or

3l = vt3 + (1/2)at3^2 = vt3 + (1/2).(v/t3).t3^2

thus,

3l = (1/2)v.t3

or

time taken in third case

t3 = 6l / v .......................(3)

.

.

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

vav = (l + 2l + 3l) / (t1 + t2 + t3)

= 6l / [(2l/v) + (2l/v) + (6l/v)] = 8l / (10l/v)

thus,

vav = 4v/5 ..........................(4)

and

the maximum speed will be

vmax = v .............................(5)

thus, the ratio will be

Vmax / Vav = v / (4v/5)

thus,

Vmax / Vav = 5/4