Question

The coordinates of a moving particle at any time t are given by \[x=\alpha {{t}^{3}}\]and\[y=\beta
{{t}^{3}}\] . The speed of the particle at time t is given by :

A: $\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}}$
B: $3t\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}}$
C: $3{{t}^{2}}\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}}$
D: ${{t}^{2}}\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}}$

Answer 1

Hint:All the quantities that we’re aware of, are divided into two categories, scalar and vector. We can define scalar quantities as those which have a definite magnitude but no direction. Whereas we can define a vector quantity as one with both magnitude and direction. Velocity is a vector quantity and speed is its corresponding scalar quantity. We can find the speed of the particle at time t by differentiating the displacement with respect to time and taking its magnitude.

Formulas used:
${{v}_{x}}=\dfrac{dx}{dt}$ and ${{v}_{y}}=\dfrac{dy}{dt}$

Complete step by step answer:
We know that velocity is a vector quantity. It is defined as the rate with which the displacement changes, where speed is the rate of change of distance of the particle. It is a scalar quantity.

We are given that \[x=\alpha {{t}^{3}}\]and \[y=\beta {{t}^{3}}\].
We can find the value of velocity by differentiating the value of displacement using the formula,
${{v}_{x}}=\dfrac{dx}{dt}$ and ${{v}_{y}}=\dfrac{dy}{dt}$
Hence,
$\begin{align}
& {{v}_{x}}=\dfrac{d(\alpha {{t}^{3}})}{dt}=3\alpha {{t}^{2}} \\
& {{v}_{y}}=\dfrac{d(\beta {{t}^{2}})}{dt}=3\beta {{t}^{2}} \\
\end{align}$
To find the resultant velocity, $R=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta }$ where R is the
magnitude between two vectors.
The angle between x axis and y axis is ${{90}^{\circ }}$
Hence, speed of the particle is
\[\begin{align}
& V=\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}}={{t}^{2}}\sqrt{{{(3\alpha )}^{2}}+{{(3\beta )}^{2}}} \\
& =3{{t}^{2}}\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}} \\
\end{align}\]

Thus, we can conclude that option C is the correct answer among the given options.

Note:
The above expression had no relation with the direction of the particle. When the direction is being considered while calculation, we have to use the formula $\alpha ={{\tan }^{-
1}}(\dfrac{{{v}_{y}}}{{{v}_{x}}})$ which gives the angle subtended by the particle.