Exam

NTA Abhyas

Year

2020

Subject

Physics

Chapter

Motion in One Dimension

NTA Abhyas 2020 · Physics · Motion in One Dimension

A particle moves according to the law x = rcos ( )/(2) . The...

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

Step-by-step solution

Given, $x = rcos \dfrac{\pit}{2}$

$⇒ v = \dfrac{dx}{dt}$

$= - \dfrac{\pi}{2}rsin \dfrac{\pit}{2}$

Graphically it can be represented as shown in figure.



Distance covered by the particle

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

$0$ to $2 s$ and $2$ to $3 s$

$= {∫}_{0}^{2}vdt + {∫}_{2}^{3}vdt$

$= \left|- {∫}_{0}^{2}\dfrac{\pi}{2}rsin \dfrac{\pi}{2}tdt\right| + \left|{∫}_{2}^{3}- \dfrac{\pi}{2}rsin \dfrac{\pi}{2}tdt\right|$

$= \left|\dfrac{- \pi}{2}r{∫}_{0}^{2}sin \dfrac{\pit}{2}dt\right| + \left|\dfrac{- \pi}{2}r{∫}_{2}^{3}sin \dfrac{\pi}{2}tdt\right|$

$= \left|{\left[\dfrac{\pir}{2}\dfrac{cos \dfrac{\pit}{2}}{\dfrac{\pi}{2}}\right]}_{0}^{2}\right| + \left|{\left[\dfrac{\pir}{2}\dfrac{cos \dfrac{\pi}{2}t}{\dfrac{\pi}{2}}\right]}_{2}^{3}\right|$

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

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