A particle's velocity v at time t is given by $v=2 e^{2 t} \cos \frac{\pi t}{3}$. The least value of t at which the acceleration becomes zero, is
Answer & explanation
Correct answer: option 3
We have,
$v =2 e^{2 t} \cos \frac{\pi t}{3}$
$\Rightarrow \frac{d v}{d t} =4 e^{2 t} \cos \frac{\pi t}{3}-\frac{2 \pi}{3} e^{2 t} \sin \frac{\pi t}{3}$
Now,
Acceleration = 0
$\Rightarrow 4 e^{2 t} \cos \frac{\pi t}{3}-\frac{2 \pi}{3} e^{2 t} \sin \frac{\pi t}{3}=0$
$\Rightarrow \tan \frac{\pi t}{3}=\frac{6}{\pi} \Rightarrow t=\frac{3}{\pi} \tan ^{-1}\left(\frac{6}{\pi}\right)$