Starting from rest a particle moves in a straight line with acceleration:
\[\] $\left\{ \begin{align} & a={{(25-{{t}^{2}})}^{\dfrac{1}{2}}}m{{s}^{-2}},0\le t\le 5s \\ & a=\dfrac{3\pi }{8}m{{s}^{-2}},t\ge 5s \\ \end{align} \right. \\ $
Find the velocity of a particle at t=7s?
A. $11\,m{{s}^{-1}}$
B. $22\,m{{s}^{-1}}$
C. $33\,m{{s}^{-1}}$
D. $44\,m{{s}^{-1}}$

Hint:Equation for acceleration is given and we know the connection between acceleration and velocity of particles. When we integrate acceleration, we get velocity and if we differentiate velocity with respect to time, we get acceleration of particles.

Formula used:
velocity, $v=\int{adt}$
Where a is the acceleration.
Also, we use the already known result in integration
That is:
\[\int{({{a}^{2}}}-{{y}^{2}}{{)}^{\dfrac{1}{2}}}dy=\dfrac{y}{2}\sqrt{{{a}^{2}}-{{y}^{2}}}+\dfrac{{{a}^{2}}}{2}{{\sin }^{-1}}\left( \dfrac{y}{a} \right)\]

Complete step by step solution:
Velocity of a particle is the rate of change of displacement and acceleration of a particle is the rate of change velocity. Or else we can say that acceleration is the time derivative of velocity or can also be expressed as the second derivative of distance with respect to time.
$velocity,v=\int{adt}$

We have to find the velocity of a particle at time t=7s. But up to 5s particles have a varying acceleration with respect to time. And after 5s acceleration of the particle becomes constant. Therefore, the interval of integration is divided into two parts. That is velocity of the particle at time, t=7s is:
$v=\int\limits_{0}^{5}{{{(25-{{t}^{2}})}^{\dfrac{1}{2}}}dt+\int\limits_{5}^{7}{\dfrac{3\pi }{8}}}dt$

On integrating first part we get:
\[\int\limits_{0}^{5}{(25}-{{t}^{2}}{{)}^{\dfrac{1}{2}}}dt=\left[ \dfrac{t}{2}\sqrt{{{5}^{2}}-{{t}^{2}}}+\dfrac{25}{2}{{\sin }^{-1}}\left( \dfrac{t}{5} \right) \right]_{0}^{5} \\
\Rightarrow \int\limits_{0}^{5}{(25}-{{t}^{2}}{{)}^{\dfrac{1}{2}}}dt =\left[ \dfrac{5}{2}\sqrt{{{5}^{2}}-{{5}^{2}}}+\dfrac{25}{2}{{\sin }^{-1}}\left( \dfrac{5}{5} \right) \right]-0 \\
\Rightarrow \int\limits_{0}^{5}{(25}-{{t}^{2}}{{)}^{\dfrac{1}{2}}}dt =\dfrac{25\pi }{4}\]

on integrating second part we get:
\[\int\limits_{5}^{7}{\dfrac{3\pi }{8}}dt=\left[ \dfrac{3\pi t}{8} \right]_{5}^{7} \\
\Rightarrow \int\limits_{5}^{7}{\dfrac{3\pi }{8}}dt =\dfrac{21\pi }{8}-\dfrac{15\pi }{8}=\dfrac{6\pi }{8} \\
\Rightarrow \int\limits_{5}^{7}{\dfrac{3\pi }{8}}dt =\dfrac{3\pi }{4}\]
Therefore, velocity of the particle at 7s is:
Velocity, $v=\dfrac{25\pi }{4}+\dfrac{3\pi }{4}=7\pi $
If we take the value of π as $\dfrac{22}{7}$ , then velocity of the particle, v=$22\,m{{s}^{-1}}$.

Hence, the correct answer is option B.

Notes: Here we are considering speed as continuous. Note that here we have to split the integral into two parts because we are asked to find the velocity at time, t=7s but acceleration of this particle changes after time, t=5s. We have to consider both the intervals for finding velocity.