Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Applications of Derivatives

Question:

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

Options:

0

$\frac{3}{2}$

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

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

Correct Answer:

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

Explanation:

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

$\Rightarrow 4 e^{2 t} \cos \frac{\pi t}{3}-\frac{2 \pi}{3} e^{2 t} \sin \frac{\pi t}{3}=0$

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