If $t = \dfrac{{{v^2}}}{2}$, then $\left( {\dfrac{{ - df}}{{dt}}} \right)$ is equal to, (where $f$ is acceleration)
A. $(1){f^2}$
B. $(2){f^3}$
C. $(3) - {f^3}$
D. $(4) - {f^2}$

Hint: In order to solve this question, first we will differentiate both the sides of the given equation. Then we will replace some value in the equation with acceleration as the answer of this question comes out to be in terms of acceleration. Then on again differentiating the new expression which is formed, we will get the answer.

Complete step by step solution:
In this question we are given that $t = \dfrac{{{v^2}}}{2}$
On cross multiplying the left hand side and the right hand side, we get,
${v^2} = 2t$
On differentiating both the sides of the above expression with respect to time, we get,
$2v\dfrac{{dv}}{{dt}} = 2$
Now, on cancelling $2$ on both the sides, we get,
$v\dfrac{{dv}}{{dt}} = 1$
$\dfrac{{dv}}{{dt}} = \dfrac{1}{v}.....(1)$
We know that $\dfrac{{dv}}{{dt}}$ is equal to the acceleration $f$, so,
$\dfrac{{dv}}{{dt}} = \dfrac{1}{v} = f$
So, on equating the last two terms of the above equation, we get,
$f = \dfrac{1}{v}$
On again differentiating both the sides with respect to time, we get,
$\dfrac{{df}}{{dt}} = - \dfrac{1}{{{v^2}}}\dfrac{{dv}}{{dt}}$
On putting the value of $\dfrac{{dv}}{{dt}}$ from equation (1), we get,
$\dfrac{{df}}{{dt}} = - \dfrac{1}{{{v^2}}}\left( {\dfrac{1}{v}} \right)$
On further solving, we get,
$\dfrac{{df}}{{dt}} = - \dfrac{1}{{{v^3}}}$
On multiplying both the sides with a negative sign, we get,
$ - \dfrac{{df}}{{dt}} = \dfrac{1}{{{v^3}}}......(2)$
We know that,
$f = \dfrac{1}{v}$
On taking cube on both the sides of the above equation, we get,
${f^3} = \dfrac{1}{{{v^3}}}$
Now, on putting the above value in equation (2), we get,
$ - \dfrac{{df}}{{dt}} = {f^3}$
So, the value of $\left( {\dfrac{{ - df}}{{dt}}} \right)$ is equal to ${f^3}$.
So, the correct answer is $(2){f^3}$.

Note:
In order to solve such questions which are based on differentiation, we must note that if a term is differentiated in terms of itself, then we will simply differentiate but if a term is differentiated in terms of another term, then we will differentiate it again in terms of the term given to us.