The equation of motion of a particle

Question:

The equation of motion of a particle is $s=2 t^{2}+\sin 2 t$, where $s$ is in metres and $t$ is in seconds. The velocity of the particle when its acceleration is $2 \mathrm{~m} / \mathrm{sec}^{2}$, is

(a) $\pi+\sqrt{3} \mathrm{~m} / \mathrm{sec}$

(b) $\frac{\pi}{3}+\sqrt{3} \mathrm{~m} / \mathrm{sec}$

(c) $\frac{2 \pi}{3}+\sqrt{3} \mathrm{~m} / \mathrm{sec}$

(d) $\frac{\pi}{3}+\frac{1}{\sqrt{3}} \mathrm{~m} / \mathrm{sec}$

Solution:

(b) $\frac{\pi}{3}+\sqrt{3} \mathrm{~m} / \mathrm{sec}$

According to the question,

$s=2 t^{2}+\sin 2 t$

$\Rightarrow \frac{d s}{d t}=4 t+2 \cos 2 t$

$\Rightarrow \frac{d^{2} s}{d t^{2}}=4-4 \sin 2 t$

$\Rightarrow 4-4 \sin 2 t=2$

$\Rightarrow 4 \sin 2 t=2$

$\Rightarrow \sin 2 t=\frac{1}{2}$

$\Rightarrow 2 t=\frac{\pi}{6}$

Now,

$\frac{d s}{d t}=4\left(\frac{\pi}{12}\right)+2 \cos \left(\frac{\pi}{6}\right)$

$\Rightarrow \frac{d s}{d t}=\frac{\pi}{3}+\sqrt{3} \mathrm{~m} / \mathrm{sec}$

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