 # If the distance ${\bf{s}}$ covered by a particle in time $t$ is proportional to the cube root of its velocity, then the acceleration is:a constant1. proportional to ${{\bf{s}}^{\bf{3}}}$2. proportional to ${\bf{1}}/{{\bf{s}}^{\bf{3}}}$3. proportional to ${{\bf{s}}^{\bf{5}}}$4. proportional to ${\bf{1}}/{{\bf{s}}^{\bf{5}}}$ Verified
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Hint: In this question we have to correlate the terms given such that the distance is proportional to the cube root of the velocity in this way. And then we can use differentiation with respect to time to solve this question.

As asked in the question that we have to calculate acceleration according to given data,
Distance ${\bf{s}}$ of the particle is proportional to the cube root of its velocity such that
$s{\rm{ }}\alpha {\rm{ }}{v^{\dfrac{1}{3}}}$

Now, writing it in simplest form we have
$v{\rm{ }}\alpha {\rm{ }}{{\rm{s}}^3}\, —-(1)$

But, we can write velocity as the function of distance and time such that
$v = \dfrac{{ds}}{{dt}}$, put this vale in equation$(1)$, we get
$\Rightarrow \dfrac{{ds}}{{dt}}\alpha {{\rm{s}}^3}$
$\Rightarrow v = \dfrac{{ds}}{{dt}} = p{{\rm{s}}^3} —-(2)$ $p$ is the proportionality constant

While acceleration is given by,
$a = \dfrac{{dv}}{{dt}} = \dfrac{d}{{dt}}\left( {p{s^3}} \right)$ , from $(2)$
$\Rightarrow a = 3p{s^2}\left( {\dfrac{{ds}}{{dt}}} \right)$ … on differentiating … from $(2)$
$\Rightarrow a = 3{p^2}{s^5}$

Thus, the answer is the acceleration is directly proportional to the fifth root of ${\bf{s}}$ i.e., Option 4.

Note: It is very easy to solve this type of question because here are the very basic concepts used from mathematics as well proportionality we are learning from early standards and are very much aware of the concepts. Along with this we must know differentiation and basic concept of distance, velocity and acceleration.