Question

A point particle moves on the x-y plane according to the law $x=a\sin (\omega t)$ and $y=a(1-\cos (\omega t))$ where $a$ and $\omega$ are positive constants and $t$ is in seconds. Find the distance covered in time ${{t}_{0}}$.
A. $a\omega {{t}_{0}}$
B. $\sqrt{2{{a}^{2}}+2{{a}^{2}}\cos (\omega {{t}_{0}})}$
C. $2a\sin \left( \dfrac{\omega {{t}_{0}}}{2} \right)$
D. $2a\cos \left( \dfrac{\omega {{t}_{0}}}{2} \right)$

Answer 1

Hint: Calculate rate of change of $x$ and $y$ with respect to time. This gives us velocity of point particles in $x$ and $y$ direction respectively. The magnitude of the velocity vector is the speed of the particle. Distance can be calculated by multiplying the speed of a particle with time period.

Formula used: Distance travelled $d=\left| \mathbf{v} \right|{{t}_{0}}$, magnitude of velocity $\left| \mathbf{v} \right|=\sqrt{v_{x}^{2}+v_{y}^{2}}$

Complete step by step answer:
We are given functions of $x$ and $y$ coordinate in terms of time $t$. The change in $x$ and $y$ coordinates is the displacement of point particles in $x$ and $y$ direction respectively.
Velocity of a particle is defined as rate of change of displacement. So the rate of change of $x$ and $y$ coordinates is the velocity of point particles along $x$ and $y$ direction respectively.
We can calculate the velocity of a particle by differentiating its functions of position. Therefore,
${{v}_{x}}=\dfrac{dx}{dt}=\dfrac{d(a\sin (\omega t))}{dt}=a\omega \cos (\omega t)$
${{v}_{y}}=\dfrac{dy}{dt}=\dfrac{d(a(1-\cos (\omega t))}{dt}=a\omega \sin (\omega t)$
Where ${{v}_{x}}$ and ${{v}_{y}}$ denote velocities along the $x$ and $y$ axis respectively.
We can calculate the magnitude of velocity by taking the square root of the sum of squares of velocities along the $x$ and $y$ axis. Mathematically, we can write
$\left| \mathbf{v} \right|=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{{{(a\omega \cos \omega t)}^{2}}+{{(a\omega \sin \omega t)}^{2}}}$
$\left| \mathbf{v} \right|=\sqrt{{{a}^{2}}{{\omega }^{2}}co{{s}^{2}}\omega t+{{a}^{2}}{{\omega }^{2}}{{\sin }^{2}}\omega t}=\sqrt{{{a}^{2}}{{\omega }^{2}}(co{{s}^{2}}\omega t+{{\sin }^{2}}\omega t)}$
We know from trigonometric identity that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. Therefore,
$\left| \mathbf{v} \right|=a\omega $
Product of speed and time period gives the distance travelled during time period. Hence, distance travelled in time ${{t}_{0}}$ is given by
$d=\left| \mathbf{v} \right|{{t}_{0}}=a\omega {{t}_{0}}$
The point particle travels distance $a\omega {{t}_{0}}$ in time ${{t}_{0}}$.

So, the correct answer is “Option A”.

Note: Distance is a scalar quantity. It is defined as the total path travelled by the particle.
Displacement of a particle is the change in its position. It is the minimum distance; a particle will have to travel from one point to another. If a particle travels and comes back to the same place then its displacement is zero. Displacement is a vector quantity.