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). \[\dfrac{3}{2}\]

Hint: Obtain the value of the mathematical expression for velocity as a function of space and as a function of time. Obtain the mathematical expression for space average velocity and time average velocity and put the above values and then take the ratio to find the answer.
Formula used:
Space average velocity = \[\dfrac{\int{vdx}}{\int{dx}}\]
Time average velocity = \[\dfrac{\int{vdt}}{\int{dt}}\]

Complete step-by-step answer:
When velocity is averaged over a time interval then it is known as time average velocity when average to the average of elected over space period then it is known as space average velocity.
Here, it is given that particles' initial velocity is zero.
In case of uniform acceleration, we can use the following formula to find the final velocity of the object.
\[v=\text{ }u\text{ }+\text{ }at\text{ }\to \text{ 1}\]
u is initial velocity and which is zero as given in question.
So, it become,
$\begin{align}
  & v=0+at \\
 & v=at \\
\end{align}$
From the equation of motion
\[x=ut+\dfrac{1}{2}a{{t}^{2}}\text{ }\to \text{ 2}\]
As u = 0
\[x=\dfrac{1}{2}a{{t}^{2}}\]
We can use this to express the velocity in terms of x.
\[\begin{align}
  & x=\dfrac{1}{2}a{{t}^{2}} \\
 & x=\dfrac{1}{2a}{{\left( at \right)}^{2}} \\
\end{align}\]
Putting $v=at$
\[\begin{align}
  & x=\dfrac{1}{2a}{{v}^{2}} \\
 & {{v}^{2}}=2ax \\
 & v=\sqrt{2ax} \\
\end{align}\]
Now substituting \[v=\sqrt{2ax}\] in space average velocity i.e.
Space average velocity =\[\dfrac{\int{vdx}}{\int{dx}}\]
 = \[\dfrac{\int{\sqrt{2ax}dx}}{x}\]
 = \[\dfrac{\sqrt{2a}\int{{{x}^{\dfrac{1}{2}}}}dx}{x}\]
= \[\sqrt{2a}\dfrac{\dfrac{{{x}^{\dfrac{1}{2}+1}}}{\dfrac{1}{2}+1}}{x}\]
= \[\dfrac{2}{3}\sqrt{2ax}\]
So, we can write that,
Space average velocity \[=\dfrac{\int{vdx}}{\int{dx}}=\dfrac{2}{3}v\]
Now substitute $v=at$ in time average velocity
= \[\dfrac{\int{vdt}}{\int{dt}}\]
 = \[\dfrac{\int{atdt}}{t}\]
 = \[\dfrac{\dfrac{a{{t}^{2}}}{2}}{t}\]
 = \[\dfrac{at}{2}\] = \[\dfrac{v}{2}\]= \[\dfrac{1}{2}v\]
 The ratio of the space average velocity and time average velocity is,
Ratio =\[\dfrac{\text{space average velocity}}{\text{time average velocity}}\]
= \[\dfrac{\dfrac{2v}{3}}{\dfrac{1v}{2}}\]
=\[\dfrac{4}{3}\]
The correct option is (C).

Note: When velocity of a particle is non uniform it may be a function of space (position) or time or a function of both. In that case these two types of parameters of velocity are externally useful in mechanics. If we are simply asked to find the average velocity then we just need to find the ratio of the change in position to the change in time.