Question

A particle moves in a straight line and its position at time t is given by ${{x}^{3}}=2t-1$. Its velocity is given by
$\begin{aligned} & \left( 1 \right)\dfrac{1}{{{x}^{2}}} \\\ & \left( 2 \right)\dfrac{2}{3{{x}^{2}}} \\\ & \left( 3 \right)\dfrac{2}{3x} \\\ & \left( 4 \right)\dfrac{1}{3x} \\\ \end{aligned}$

Answer 1

Here the equation for a particle moving in a straight line is given. Hence, differentiating the given equation with respect to time we get a ratio of displacement to time which is known as the velocity. Hence, the velocity of a particle can be described as the rate of change of displacement of a body with respect to time.

Complete answer:
Given that,
${{x}^{3}}=2t-1$
Now differentiating the given equation with respect to time we get,
$3x\dfrac{dx}{dt}=2$
Then by rearranging the equation the equation becomes,
$\dfrac{dx}{dt}=\dfrac{2}{3}x$ …………….(1)
We know that the velocity is the ratio of displacement to time.
That is,
$v=\dfrac{dx}{dt}$ ………..(2)
Hence, by equating equation (1) and (2) we get,
$v=\dfrac{dx}{dt}=\dfrac{2}{3}x$
Hence, the velocity of a particle can be described as the rate of change of displacement of a body with respect to time.
Therefore, the velocity of the given particle travelling in a straight line is $\dfrac{2}{3}x$.

Thus, option (3) is correct.

Additional information:
The velocity of a particle can otherwise be described as the change
in its position while considering a frame of reference. And also the velocity will always depend on time, so that it is a function of time.

Note:
The velocity of a particle can be described as the rate of change of displacement of a body with respect to time. If we want to calculate the acceleration of a particle, we have to again differentiate the velocity with respect to time. Thus, the acceleration can be defined as the rate of change of velocity of a particle with respect to time.