Physics - Kinematics Question with Solution | TestHub

Given,$x=r\cos \frac{\pi t}{2}$

$\Rightarrow v=\frac{dx}{dt}$

$=-\frac{\pi }{2}r\sin \frac{\pi t}{2}$

Graphically it can be represented as shown in figure.



Distance covered by the particle

$=$Magnitude of area of$x-t$curve between

$0$to$\mathrm{2\; s}$and$2$to$\mathrm{3\; s}$

$={\int }_0^{2}vdt+{\int }_2^{3}vdt$

$=\left|-{\int }_0^2\frac{\pi }{2}r\sin \frac{\pi }{2}tdt\right|+\left|{\int }_2^3-\frac{\pi }{2}r\sin \frac{\pi }{2}tdt\right|$

$=\left|\frac{-\pi }{2}r{\int }_0^2\sin \frac{\pi t}{2}dt\right|+\left|\frac{-\pi }{2}r{\int }_2^3\sin \frac{\pi }{2}tdt\right|$

$=\left|{\left[\frac{\pi r}{2}\frac{\cos \frac{\pi t}{2}}{\frac{\pi }{2}}\right]}_0^2\right|+\left|{\left[\frac{\pi r}{2}\frac{\cos \frac{\pi }{2}t}{\frac{\pi }{2}}\right]}_2^3\right|$

$=\left|r\left(\cos \pi -\cos 0\right)\right|+\left|r\left(\cos \frac{3\pi }{2}-\cos \pi \right)\right|$

$=r\left|-1-1\right|+r\left|0-\left(-1\right)\right|=2r+r=3r$