Question

A lift going upward perform the first part of ascent with uniform acceleration a and then remaining part with uniform retardation 2a .if total time of ascent is t the depth of the lift shaft is

Answer 1

Hii Dear,

◆ Answer -

h = at^2/3

◆ Explanation -

Let t1 and t2 be time taken during accelerated motion amd retarded motion respectively.

During acceleration,

a = (v - 0) / t1

t1 = v/a

During retardation,

-2a = (0 - v) / t2

t2 = v/2a

Total time of ascent is -

t = t1 + t2

t = v/a + v/2a

t = 3v/2a

Putting v = 2h/t,

t = 3/2a × 2h/t

h = t^2 × a / 3

h = at^2/3

Therefore, depth of the shaft will be at^2/3.

Thanks dear...