Question

If the distance $s$ covered by a particle in time $t$ is proportional to the cube root of its velocity, then the acceleration is

• A. a constant
• B. proportional to $s^3$
• C. proportional to $\frac{1}{s^3}$
• D. proportional to $s^5$
• E. proportional to $\frac{1}{s^5}$

Answer 1

The correct option is D

proportional to $s^5$

Step 1: Given Data

$$\begin{gathered} s\propto v^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \\ {s}^3\propto v \end{gathered}$$

The distance covered by the particle is $s$.

The velocity of the particle is $v$.

Step 2: Formula used

We can write $v$ as the change in distance with respect to change in time,

$v=\frac{ds}{dt}$

We can also write acceleration $a$ in terms of change in velocity with respect to change in time,

$a=\frac{dv}{dt}$

Step 3: Evaluate the acceleration

$$\begin{gathered} \frac{ds}{dt}\propto s^3 \\ v=k{s}^3\mathrm{Where}\mathrm{k}\mathrm{is}\mathrm{constant}.....\left(1\right) \\ a=\frac{dv}{dt}=\frac{d\left[k{s}^3\right]}{dt} \\ a=k\times 3s^2\left(\frac{ds}{dt}\right) \\ a=3k{s}^2\times k{s}^3\mathrm{From}\mathrm{equation}\left(1\right) \\ a=3k^2{s}^5 \\ a\propto s^5(\mathrm{As}3k^2\mathrm{is}\mathrm{a}\mathrm{constant}) \end{gathered}$$

Thus, the acceleration of a particle is proportional to $s^5$

Hence, option D is the correct option.