Question

What is the distance travelled by a particle, during the time ‘t’ if the particle moves in x, y plane according to the following equation. Also, find its velocity.
$\begin{align}
  & \Rightarrow x=a\sin (\omega t) \\
 & \Rightarrow y=a[1-\cos (\omega t)] \\
\end{align}$

Answer 1

Hint: In order to calculate the distance travelled, we need to find the speed of the particle. This can be done by finding out the X and Y components of velocity and then finding the magnitude of velocity. This will give us the required speed of the particle which when multiplied by time will give us the total distance travelled.

Complete step-by-step answer:
To find the velocity of a particle as its different components in X and Y, we shall differentiate its displacement along the X and Y direction with respect to time. This can be done as follows:
$\Rightarrow {{v}_{x}}=\dfrac{dx}{dt}$
Where, $x=a\sin (\omega t)$
Putting it in the above velocity equation, we get:
$\begin{align}
  & \Rightarrow {{v}_{x}}=\dfrac{d[a\sin (\omega t)]}{dt} \\
 & \Rightarrow {{v}_{x}}=a\omega \cos (\omega t) \\
\end{align}$
And, for the Y-component, we have:
$\Rightarrow {{v}_{y}}=\dfrac{dy}{dt}$
Where, $y=a-a\cos (\omega t)$
Putting it in the above velocity equation, we get:
$\begin{align}
  & \Rightarrow {{v}_{y}}=\dfrac{d[a-a\cos (\omega t)]}{dt} \\
 & \Rightarrow {{v}_{y}}=-a\omega [-\sin (\omega t)] \\
 & \Rightarrow {{v}_{y}}=a\omega \sin (\omega t) \\
\end{align}$
Hence, the velocity of the particle comes out to be $[a\omega \cos (\omega t)\widehat{i}+a\omega \sin (\omega t)\widehat{j]}$ .
Now, the speed of particle can be given by:
$\Rightarrow v=\sqrt{v_{x}^{2}+v_{y}^{2}}$
Which is equal to:
$\begin{align}
  & \Rightarrow v=\sqrt{{{[a\omega \cos (\omega t)]}^{2}}+{{[a\omega \sin (\omega t)]}^{2}}} \\
 & \Rightarrow v=\sqrt{{{a}^{2}}{{\omega }^{2}}[{{\cos }^{2}}(\omega t)+{{\sin }^{2}}(\omega t)]} \\
 & \therefore v=a\omega \\
\end{align}$
Hence, the speed of the particle comes out to be $a\omega $ .
Therefore, the distance travelled (say d) is equal to:
$\begin{align}
  & \Rightarrow d=speed\times time \\
 & \Rightarrow d=a\omega \times t \\
 & \therefore d=a\omega t \\
\end{align}$
Hence, the distance travelled by the particle, during the time ‘t’ comes out to be $a\omega t$ .

Note: We cannot simply add the squares of the given expression and find its root for displacement in X and Y direction as they are not some constant values. The particle is moving on a complex path of its own. So, the best way to find its distance travelled is by calculating the speed of the particle and then multiplying it with time, so that we can simply get the distance travelled.