A body moves for a total of nine second starting from rest with uniform acceleration and then with uniform retardation, which is twice the value of acceleration and then stop. The duration of uniform acceleration is

Options

A. 3s
B. 4.5s
C. 5s
D. 6s

Answer

D

Explanation

Acceleration = a, Retardation = 2a
Time taken to reach maximum velocity
`t_1 = v/a`
Time taken to stop after reaching maximum velocity
`t_2 = (v)/(2a)`

Now, `t_1 + t_2 = 9 rArr v/a + (v)/(2a) = 9`
rArr v/a = 6`
duration of acceleration `= t_1 = v/a = 6sec`