Question

If a particle moves such that the displacement (s) is proportional to the square of the velocity (v), then its acceleration (a) is

• A. proportional to \(s^2\)
• B. proportional to \(1/s\)
• C. proportional to \(1/s^2\)
• D. a constant

Hint:

Write s = c v^2, differentiate with respect to time and use ds/dt = v.

Answer 1

The Correct Option is D

Step 1: Compare with a known motion:
Uniform acceleration from rest gives $v^2 = 2as$, so $s = \dfrac{v^2}{2a}$, which is exactly the form $s \propto v^2$.

Step 2: Do the algebra the other way:
From $s = cv^2$, use $a = v\dfrac{dv}{ds}$. Now $\dfrac{ds}{dv} = 2cv$, so $\dfrac{dv}{ds} = \dfrac{1}{2cv}$.
$a = v\cdot\dfrac{1}{2cv} = \dfrac{1}{2c}$.

Step 3: Conclusion:
No $s$ remains in the result, so $a$ is constant.

Final Answer:
Option (D). \[ \boxed{\text{a constant (D)}} \]