A particle moving along x-axis has acceleration `f`, at time `t`, given by `f = f_0 (1 - (t)/(T))`, where `f_0` and `T` are constant.
The particle at `t = 0` has zero velocity. In the time interval between `t = 0` and the instant when `f = 0`, the particle's velocity `(v_x)` is :

Options

A. `f_(0)T`
B. `(1)/(2)f_(0)T^(2)`
C. `f_(0)T^(2)`
D. `(1)/(2)f_(0)T`

Answer

D

Explanation

Acceleration,
`f=f_(0)(1-(t)/(T))`
`implies (dv)/(dt)=f_(0)(1-(t)/(T))`
`intdv=int_(0)^(T)f_(0)(1-(t)/(T))dt`
`impliesv=f_(0)[t-(t^(2))/(2T)]_(0)^(T)=f_(0)[T-(T^(2))/(2T)]`
`v=f_(0)(T-(T)/(2)),v=f_(0)"(T)/(2)`