Question

A particle moves according to the law $x=rcos\frac{\pi t}{2}$ . The distance covered by it in the time interval between $t=0$ and $t=3s$ is $nr$ . What is the value of $n$ ?

Answer 1

Answer: 3

Given, $x=rcos\frac{\pi t}{2}$
$\Rightarrow v=\frac{dx}{dt}$
$=-\frac{\pi }{2}rsin\frac{\pi t}{2}$
Graphically it can be represented as shown in figure.

Distance covered by the particle
$=$ Magnitude of area of $x-t$ curve between
$0$ to $2s$ and $2$ to $3s$
$={\int }_0^{2}vdt+{\int }_2^{3}vdt$
$=\left|-{\int }_0^2\frac{\pi }{2}rsin\frac{\pi }{2}tdt\right|+\left|{\int }_2^3-\frac{\pi }{2}rsin\frac{\pi }{2}tdt\right|$
$=\left|\frac{-\pi }{2}r{\int }_0^{2}sin\frac{\pi t}{2}dt\right|+\left|\frac{-\pi }{2}r{\int }_2^{3}sin\frac{\pi }{2}tdt\right|$
$=\left|{\left[\frac{\pi r}{2}\frac{cos\frac{\pi t}{2}}{\frac{\pi }{2}}\right]}_0^2\right|+\left|{\left[\frac{\pi r}{2}\frac{cos\frac{\pi }{2}t}{\frac{\pi }{2}}\right]}_2^3\right|$
$=\left|r\left(cos\pi -cos0\right)\right|+\left|r\left(cos\frac{3\pi }{2}-cos\pi \right)\right|$
$=r\left|-1-1\right|+r\left|0-\left(-1\right)\right|=2r+r=3r$