Starting from rest a particle moves in a straight line with acceleration $$a = {\left( {25 - {t^2}} \right)^{\frac{1}{2}}}m/{s^2}$$     for $$0 \leqslant t \leqslant 5\,s,a = \frac{{3\pi }}{8}m/{s^2}$$      for $$t > 5\,s.$$   The velocity of particle at $$t = 7\,s$$  is:

Solution :
$${V_7} = {V_5} + \frac{{3\pi }}{8} \times 2$$
To calculate $${V_5}$$ i.e. velocity at end of $$5\sec .$$
$$a = {\left( {25 - {t^2}} \right)^{\frac{1}{2}}} \Rightarrow \frac{{dv}}{{dt}} = {\left( {25 - {t^2}} \right)^{\frac{1}{2}}}$$
\[\int {dv} = \int 5 {\left( {25 - 25{{\sin }^2}\theta } \right)^{\frac{1}{2}}}\cos \theta d\theta \left\{ {\begin{array}{*{20}{l}} {t = 5\sin \theta }\\ {\frac{{dt}}{{d\theta }} = 5\cos \theta }\\ {t = 0,\theta = 0}\\ {t = 5,\theta = \frac{\pi }{2}} \end{array}} \right.\]
$$\eqalign{ & = \int_0^{\frac{\pi }{2}} 5 \times 5{\cos ^2}\theta d\theta = \frac{{25\pi }}{4} \cr & {V_7} = \frac{{25\pi }}{4} + \frac{{3\pi }}{4} = \frac{{28\pi }}{4} = 7\pi = 22\,m/s \cr} $$