Question

The position coordinates of a particle moving in a 3-D coordinate system is given by
$\begin{gathered} x = a\cos \omega t \\\ y = a\sin \omega t \\\ \end{gathered}$
and $z = a\omega t$

The speed of the particle is:
$\begin{gathered} {\text{A}}{\text{. }}a\omega \\\ {\text{B}}{\text{. }}\sqrt 3 a\omega \\\ {\text{C}}{\text{. }}\sqrt 2 a\omega \\\ {\text{D}}{\text{. }}2a\omega \\\ \end{gathered}$

Answer 1

Hint: We are given the position coordinates of the particle in three dimensions. The velocity can be directly calculated by taking the time derivative of the co-ordinates. The speed of a particle is equal to the magnitude of velocity of the particle.

Complete step-by-step answer:
We are given the following values of the co-ordinates x, y and z.
$\begin{gathered} x = a\cos \omega t \\\ y = a\sin \omega t \\\ z = a\omega t \\\ \end{gathered}$

We notice that these are functions of time and a and $\omega$ are constants. The velocity can be calculated by taking the time derivative of the these position coordinates as follows:
The x-component of velocity is equal to the time derivative of x-coordinate.
${{\text{v}}_x} = \dfrac{{dx}}{{dt}}$

Putting the expression for x, we get
$\begin{gathered} {{\text{v}}_x} = \dfrac{d}{{dt}}\left( {a\cos \omega t} \right) \\\ \Rightarrow {{\text{v}}_x} = a\dfrac{d}{{dt}}\left( {\cos \omega t} \right) \\\ \Rightarrow {{\text{v}}_x} = - a\omega \sin \omega t \\\ \end{gathered}$

Similarly, we can calculate the other components of velocity.
$\begin{gathered} {{\text{v}}_y} = \dfrac{d}{{dt}}\left( {a\sin \omega t} \right) = a\dfrac{d}{{dt}}\left( {\sin \omega t} \right) = a\omega \cos \omega t \\\ {{\text{v}}_z} = \dfrac{d}{{dt}}\left( {a\omega t} \right) = a\omega \\\ \end{gathered}$

Now from these components, we can calculate the magnitude of velocity as follows:
${\text{v}} = \sqrt {{\text{v}}_x^2 + {\text{v}}_y^2 + {\text{v}}_z^2}$

Substituting various expressions, we can get the required value of speed as follows:
$\begin{gathered} {\text{v}} = \sqrt {{{\left( { - a\omega \sin \omega t} \right)}^2} + {{\left( {a\omega \cos \omega t} \right)}^2} + {{\left( {a\omega } \right)}^2}} \\\ \Rightarrow {\text{v}} = \sqrt {{a^2}{\omega ^2}{{\sin }^2}\omega t + {a^2}{\omega ^2}{{\cos }^2}\omega t + {a^2}{\omega ^2}} \\\ \Rightarrow {\text{v}} = \sqrt {{a^2}{\omega ^2}\left( {{{\sin }^2}\omega t + {{\cos }^2}\omega t} \right) + {a^2}{\omega ^2}} \\\ \end{gathered}$

Using the identity, $\left( {{{\sin }^2}\omega t + {{\cos }^2}\omega t} \right) = 1$, we get
$\begin{gathered} {\text{v}} = \sqrt {{a^2}{\omega ^2} + {a^2}{\omega ^2}} \\\ \Rightarrow {\text{v}} = \sqrt {2{a^2}{\omega ^2}} \\\ \Rightarrow {\text{v}} = \sqrt 2 a\omega \\\ \end{gathered}$

This is the required answer so the correct answer is option C.

Note: Speed of a particle is scalar quantity while the velocity of a particle is a vector quantity. The value of speed of a particle is equal to the magnitude of the velocity. The velocity can be expressed as a vector in terms of its components in the following way.
$\overrightarrow V = {V_x}\widehat i + {V_y}\widehat j + {V_z}\widehat k$