A particle moving along X-axis has acceleration $f$, at time t, given $f=f_0\left(1-\frac{t}{T}\right)$ where $f_0$ and T are constants. The particle at t = 0 has zero velocity. When f = 0, the particle's velocity (v_x ) is

Acceleration, $f=f_0\left(1-\frac{t}{T}\right)$

                      $\frac{dv}{dt}=f_0·\left(1-\frac{t}{T}\right) \left(\because f=\frac{dv}{dt}\right)$

$\Rightarrow$                  $dv=f_0·\left(1-\frac{t}{T}\right)dt\Rightarrow \int dv=\int f_0\left(1-\frac{t}{T}\right)dt$ 

                         $v=f_{0}t-f_0\frac{t^2}{2T}+C$                                       …(i)

where, C is constant.
When t = 0, v = 0 thus from Eq. (i), C = 0

                 $v=f_{0}t-\frac{f_0}{T}·\frac{t^2}{2}$                                                    …(ii)

As,             $f=f_0\left(1-\frac{t}{T}\right)$

When        $f = 0$

$\therefore f_0\left(1-\frac{t}{T}\right)=0; f_0\ne 0\Rightarrow t=T$

Putting t = T in Eq. (ii), we get

            $v=f_{0}T-\frac{f_0}{T}·\frac{T^2}{2}=\frac{f_{0}T}{2}$