Question

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

• A. $11m/s$
• B. $22m/s$
• C. $33m/s$
• D. $44m/s$

Answer 1

Answer: (B)

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