A particle moving along $$x$$-axis has acceleration $$f$$ at time $$t,$$ given by $$f = {f_0}\left( {1 - \frac{t}{T}} \right),$$    where $${f_0}$$ and $$T$$ are constants. 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 $$\left( {{v_x}} \right)$$  is

Solution :
$$\eqalign{ & {\text{Acceleration}}\,f = {f_0}\left( {1 - \frac{t}{T}} \right) \cr & {\text{or}}\,f = \frac{{dv}}{{dt}} = {f_0}\left( {1 - \frac{t}{T}} \right)\,\,\,\left[ {\because f = \frac{{dv}}{{dt}}} \right] \cr & {\text{or}}\,dv = {f_0}\left( {1 - \frac{t}{T}} \right)dt......\left( {\text{i}} \right) \cr} $$
Integrating Eq. (i) on both sides,
$$\eqalign{ & \int d v = \int {{f_0}} \left( {1 - \frac{t}{T}} \right)dt \cr & \therefore v = {f_0}t - \frac{{{f_0}}}{T} \cdot \frac{{{t^2}}}{2} + c\,......\left( {{\text{ii}}} \right) \cr} $$
where, $$c$$ is constant of integration.
Now, when $$t = 0,v = 0$$
So, from Eq. (ii), we get $$c = 0$$
$$\eqalign{ & \therefore v = {f_0}t - \frac{{{f_0}}}{T} \cdot \frac{{{t^2}}}{2}\,......\left( {{\text{iii}}} \right) \cr & {\text{As,}}\,f = {f_0}\left( {1 - \frac{t}{T}} \right) \cr & {\text{When,}}\,f = 0 \cr & 0 = {f_0}\left( {1 - \frac{t}{T}} \right) \cr & {\text{As,}}\,{f_0} \ne 0,\,{\text{so,}}\,1 - \frac{t}{T} = 0 \cr & \therefore t = T \cr} $$
Substituting, $$t = T$$  in Eq. (iii),
we get $${v_x} = {f_0}T - \frac{{{f_0}}}{T} \cdot \frac{{{T^2}}}{2} = {f_0}T - \frac{{{f_0}T}}{2} = \frac{1}{2}{f_0}T$$