Question

A particle starts 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 average velocity of an object is the ratio of total displacement by the total time taken by the object. It is the rate at which an object changes its position from one place to another.Space average velocity is the average velocity in the space domain, i.e. Vdx is the integrated over space domain divided by the integration of dx over space domain.
Space-average velocity$= \dfrac{{\int {vdx} }}{{\int {dx} }}$
Time-average velocity$= \dfrac{{\int {vdt} }}{{\int {dt} }}$
Acceleration is a vector quantity defined as the rate of change of velocity; an object accelerates when its velocity changes are given as
$\vec a = \dfrac{{v - {v_0}}}{t} = \dfrac{{\vartriangle v}}{{\vartriangle t}}$
Here, in the question, we need to determine the ratio of space average velocity to the time-average velocity for which we will use different equations of motion to develop the expression for the space average velocity and time-averaged first by using $v = u + at$ and then, integrating them over space and time respectively. Divide the resultant equation to get the ratio.

Complete step by step answer:
Given that particle is initially at rest$u = 0m/s$, and acceleration is constant$a = a\left( {m/{s^2}} \right)$
We know ${v^2} - {u^2} = 2ax - - - (i)$ which can be written as
$v = \sqrt {2ax}$
Also

$$v = u + at - - - (ii) \\\ \Rightarrow v = 0 + at \\\ \Rightarrow v = at \\\$$

Given object starts from the point ${x_0} = 0$and reaches to the point${x_1} = x$; hence we can write
Space Average velocity from the point${x_0} = 0$to ${x_1} = x$ by using the formula $\dfrac{{\int {vdx} }}{{\int {dx} }}$, we can write

$$\dfrac{{\int\limits_{{x_0}}^{{x_1}} {vdx} }}{{\int\limits_{{x_0}}^{{x_1}} {dx} }} = \dfrac{{\int\limits_{{x_0}}^{{x_1}} {\sqrt {2ax} dx} }}{{\left( {{x_1} - {x_0}} \right)}} \\\ \Rightarrow = \dfrac{{\sqrt {2a} \int\limits_0^x {{x^{\left( {\dfrac{1}{2}} \right)}}dx} }}{{\left( {x - 0} \right)}} \\\ \Rightarrow = \dfrac{{\sqrt {2a} }}{x} \times \left( {\dfrac{{{x^{\left( {\dfrac{3}{2}} \right)}}}}{{\left( {\dfrac{3}{2}} \right)}}} \right)_0^x \\\ \Rightarrow = \dfrac{2}{3} \times \dfrac{{\sqrt {2a} }}{x} \times \left( {\sqrt x \times x} \right) \\\ \Rightarrow = \dfrac{2}{3} \times \sqrt {2ax} \\\ \Rightarrow = \dfrac{2}{3}v \\\$$

Time Average velocity from the point ${t_0} = 0$to ${t_1} = t$ by using the formula$= \dfrac{{\int {vdt} }}{{\int {dt} }}$, we can write

$$\dfrac{{\int\limits_{{t_0}}^{{t_1}} {vdt} }}{{\int\limits_{{t_0}}^{{t_1}} {dt} }}= \dfrac{{\int\limits_{{t_0}}^{{t_1}} {atdt} }}{{\left( {{t_1} - {t_0}} \right)}} \\\ \Rightarrow = \dfrac{{a\int\limits_0^t {tdt} }}{{\left( {t - 0} \right)}} \\\ \Rightarrow = \dfrac{a}{t} \times \left( {\dfrac{{{t^2}}}{2}} \right)_0^t \\\ \Rightarrow = \dfrac{1}{2} \times \dfrac{{a{t^2}}}{t} \\\ \Rightarrow = \dfrac{{at}}{2} \\\ \Rightarrow = \dfrac{1}{2}v \\\$$

Hence, the ratio of space average velocity to the time-average velocity
$\Rightarrow \dfrac{{\left( {\dfrac{{\int {vdx} }}{{\int {dx} }}} \right)}}{{\left( {\dfrac{{\int {vdt} }}{{\int {dt} }}} \right)}} = \dfrac{{\left( {\dfrac{2}{3}v} \right)}}{{\left( {\dfrac{1}{2}v} \right)}} = \dfrac{2}{3} \times \dfrac{2}{1} = \dfrac{4}{3}$
Hence,option (C) is correct.

Note: It is very important here to note that, if the displacement of the body is zero then, we cannot judge anything about the path followed by the body whereas if the distance travelled by the body is zero then, we can assure that the body is at rest and does not moved at all. The velocity of the body is associated with the displacement of the body.