Question

If the velocity of a particle moving on a straight line is proportional to the cube root of its displacement, then its acceleration is

• A. constant
• B. inversely proportional to its velocity
• C. proportional to its velocity
• D. proportional to its displacement

Hint:

If acceleration involves a derivative of velocity, expressing everything in terms of a single variable helps determine the relationship.

Answer 1

The Correct Option is B

Let \( v \propto x^{1/3} \Rightarrow v = kx^{1/3} \).
Acceleration \( a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{d}{dx}(kx^{1/3}) \cdot v \).
\[ \frac{dv}{dx} = \frac{k}{3}x^{-2/3} \Rightarrow a = \frac{k}{3}x^{-2/3} \cdot kx^{1/3} = \frac{k^2}{3}x^{-1/3} \] \[ \text{But } v = kx^{1/3} \Rightarrow x^{1/3} = \frac{v}{k} \Rightarrow x^{-1/3} = \frac{k}{v} \Rightarrow a = \frac{k^2}{3} \cdot \frac{k}{v} = \frac{k^3}{3v} \] Hence, \( a \propto \frac{1}{v} \).