Question

Starting from rest a particle moves in a straight line, whose acceleration varies with time as
$a=(29-t^2{)}^{\frac{1}{2}}{\text{m/s}}^2$ for $0\le t\le 5s$
$a=\frac{3\pi }{8}{\text{m/s}}^2$ for $t>5s$
The velocity of the particle at $t=7s$ is :

• A. $11\text{m/s}$
• B. $22\text{m/s}$
• C. $33\text{m/s}$
• D. $44\text{m/s}$

Answer 1

The correct option is B $22\text{m/s}$

$a=\frac{dv}{dt}$
$dv=adt$
$\underset{0}{\overset{v}{\int }}dv=\underset{t=0}{\overset{t=7}{\int }}adt$
As acceleration is different between $0\le t\le 5$ and $t>5$ second hence dividing the limits as t from 0 to 5 sec and from t = 5 to 7 sec:
$\underset{0}{\overset{v}{\int }}=\underset{0}{\overset{5}{\int }}[25-t^2{]}^{\frac{1}{2}}dt+\underset{5}{\overset{7}{\int }}\left[\frac{3\pi }{8}\right]dt$
$v={[\frac{t}{2}\sqrt{25-t^2}+\frac{25}{2}{\sin }^{-1}\left[\frac{t}{5}\right]]}_0^5+{\left[\frac{3\pi }{8}t\right]}_5^7$
On solving:
$v=[\frac{25}{2}.\frac{\pi }{2}]+\frac{3\pi }{8}[7-5]$
$v=\frac{25\pi }{4}+\frac{3\pi }{4}$
$v=\frac{28\pi }{4}=7\pi$
$v=22\text{m/sec}$