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

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$\frac{3}{2}$

$\frac{3}{π} \tan^{- 1} (\frac{6}{π})$

$\frac{3}{π} \cot^{- 1} (\frac{6}{π})$

answer is C.

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Detailed Solution

We have,

$\begin{array}{l} v = 2 e^{2 t} \cos \frac{π t}{3} \\ \Rightarrow \frac{d v}{d t} = 4 e^{2 t} \cos \frac{π t}{3} - \frac{2 π}{3} e^{2 t} \sin \frac{π t}{3} \end{array}$

Now,

Acceleration = 0

$\begin{aligned} = 4 e^{2 t} \cos \frac{π t}{3} - \frac{2 π}{3} e^{2 t} \sin \frac{π t}{3} = 0 \\ = \tan \frac{π t}{3} = \frac{6}{π} \Rightarrow t = \frac{3}{π} \tan^{- 1} (\frac{6}{π}) \end{aligned}$