Question

A particle's velocity $v$ at time $t$ is given by $v=2e^{2t}\cos \frac{\pi t}{3}$. The least value of $t$ at which the acceleration becomes zero is

• A. $0$
• B. $\frac{3}{2}$
• C. $\frac{3}{\pi }$ tan${}^{-1}\left(\frac{6}{\pi }\right)$
• D. $\frac{3}{\pi }$ cot${}^{-1}\left(\frac{6}{\pi }\right)$

Answer 1

Answer: (C)

$\frac{3}{\pi }$ tan${}^{-1}\left(\frac{6}{\pi }\right)$

$v=2e^{2t}\cos \frac{\pi t}{3}$
Acceleration $a=\frac{dv}{dt}=2e^{2t}[-\frac{\pi }{3}\sin \frac{\pi t}{3}+2\cos \frac{\pi t}{3}]$
$\Rightarrow -\frac{\pi }{3}\sin \frac{\pi t}{3}+2\cos \frac{\pi t}{3}=0$
$\Rightarrow \tan \frac{\pi t}{3}=\frac{6}{\pi }$
$\Rightarrow t=\frac{3}{\pi }{\tan }^{-1}\left(\frac{6}{\pi }\right)$