Question

A particle from rest with constant acceleration. The ratio of space average velocity to the time average velocity is:
$A.\,\dfrac{1}{2}$
$B.\,\dfrac{3}{4}$
$C.\,\dfrac{4}{3}$
$D.\,\dfrac{3}{2}$

Answer 1

The given question is based on the concept of space average velocity and the time average velocity. We will make use of the formulae used to calculate the space average velocity and the time average velocity. Then, we will divide these values to find the value of the ratio.

Formula used:

& {{v}_{time}}=\dfrac{\int{v\,dt}}{\int{dt}} \\\ & {{v}_{space}}=\dfrac{\int{v\,ds}}{\int{ds}} \\\ \end{aligned}$$ **Complete step by step answer:** Firstly, we will define the terms, the space average velocity and the time average velocity. Then, we will proceed with the calculation part. Space average velocity – the integration of the velocity over the space is called space average velocity. The mathematical representation of the same is given as follows. $${{v}_{space}}=\dfrac{\int{v\,ds}}{\int{ds}}$$ Where v is the velocity and ds is the change in the space with respect to the velocity. The time average velocity – the integration of the velocity over time is called the time average velocity. The mathematical representation of the same is given as follows. $${{v}_{time}}=\dfrac{\int{v\,dt}}{\int{dt}}$$ Where v is the velocity and dt is the change in the time with respect to the velocity. Now, consider the equations one by one and then proceed with the substitution. So, we have, The space average velocity, $${{v}_{space}}=\dfrac{\int{v\,ds}}{\int{ds}}$$ We will represent the velocity in terms of acceleration as follows. $$\begin{aligned} & {{v}_{space}}=\dfrac{\int{v\,\dfrac{ds}{dt}dt}}{\int{\dfrac{ds}{dt}dt}} \\\ & {{v}_{space}}=\dfrac{\int{v\,(v)dt}}{\int{vdt}} \\\ & \Rightarrow {{v}_{space}}=\dfrac{\int{{{a}^{2}}{{t}^{2}}dt}}{\int{atdt}} \\\ \end{aligned}$$ Continue further calculation by integrating the above equation. $$\begin{aligned} & {{v}_{space}}=a\dfrac{\dfrac{{{T}^{3}}}{3}}{\dfrac{{{T}^{2}}}{2}} \\\ & \Rightarrow {{v}_{space}}=\dfrac{2}{3}aT \\\ \end{aligned}$$…… (1) The time average velocity, $${{v}_{time}}=\dfrac{\int{v\,dt}}{\int{dt}}$$ We will represent the velocity in terms of acceleration as follows. $${{v}_{time}}=\dfrac{\int{atdt}}{\int{dt}}$$ Continue further calculation by integrating the above equation. $$\begin{aligned} & {{v}_{time}}=a\dfrac{{}^{{{T}^{2}}}/{}_{2}}{T} \\\ & \Rightarrow {{v}_{time}}=\dfrac{1}{2}aT \\\ \end{aligned}$$…… (2) Divide the equations (1) and (2) to find the ratio of space average velocity to the time average velocity. $$\begin{aligned} & \dfrac{{{v}_{space}}}{{{v}_{time}}}=\dfrac{\dfrac{2}{3}aT}{\dfrac{1}{2}aT} \\\ & \Rightarrow \dfrac{{{v}_{space}}}{{{v}_{time}}}=\dfrac{4}{3} \\\ \end{aligned}$$ $$\therefore $$ The ratio of space average velocity to the time average velocity is $$\dfrac{4}{3}$$ **So, the correct answer is “Option C”.** **Note:** This is a direct mathematical problem involving the integration. The mathematical representation of the space average velocity and the time average velocity should be known to solve this problem.