A lift performs the first part of its ascent with uniform acceleration a and the remaining with uniform retardation `2a.` If t is the time of ascent, find the depth of the shaft.

Options

A. `(at^2)/4`
B. `(at^2)/3`
C. `(at^2)/2`
D. `(at^2)/8`

Answer

B

Explanation

Retardation is double. Therefore retardation time
will be half.
Let `t_0`=acceleration time
Then `(t_0)/2`=retardation time
`t_0+(t_0)/2=t`
`:. t_0=(2t)/3`
Now, `s=s_1+s_2`
`=(1/2)(a)(t_0)^2+(1/2)(2a)((t_0)/2)^2`
`=3/4 at_0^2`
`=(3/4)(a)((2t)/3)^2`
`=(at^2)/3`