# The lift maintains this constant speed of V ms–1 for 25 s before decelerating uniformly to rest. The total time for descent is 40 s. Calculate the total distance that the lift descends.

Let `t_1=25 s` the time lift maintains constant speed `V` and

`t_2 =40-25 =15 s` the time lift descends decelerating uniformly.

The relation between deceleration `a` , time `t_2` and initial speed `V` is

`0 =V+a*t_2` , thus `a =-V/t_2=-V/15`

The distance that the lift travels with the same speed `V` is

`d_1 =V*t_1 =25*V`

The distance that the lift travels with uniform deceleration `a` is

`d_2 =v*t_2 +a*t_2^2/2 =V*t_2 -(V/t_2)*(t_2^2/2) =V*t_2-V*t_2/2 =(15/2)*V`

Therefore the total distance the lift travels (descends) is

`d = d_1+d_2 =25*V +15/2*V =(65/2)*V =32.5 V "meters"`

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