Q85 / 150 · MHT CET 2026 (18 April Shift 2) · Mathematics
If a particle moves such that the displacement (s) is proportional to the square of the velocity (v), then its acceleration (a) is
- Aproportional to $s^2$
- Bproportional to $1/s$
- Cproportional to $1/s^2$
- Correct answerDa constant
Solution
Given $s \propto v^2$
$s = k v^2$ where $k$ is a constant.
Differentiating both sides with respect to time $t$:
$\dfrac{ds}{dt} = 2kv \dfrac{dv}{dt}$
Since $\dfrac{ds}{dt} = v$ and $\dfrac{dv}{dt} = a$:
$v = 2kv a$
$a = \dfrac{1}{2k}$
Since $k$ is a constant, acceleration $a$ is a constant.
Answer: a constant
$s = k v^2$ where $k$ is a constant.
Differentiating both sides with respect to time $t$:
$\dfrac{ds}{dt} = 2kv \dfrac{dv}{dt}$
Since $\dfrac{ds}{dt} = v$ and $\dfrac{dv}{dt} = a$:
$v = 2kv a$
$a = \dfrac{1}{2k}$
Since $k$ is a constant, acceleration $a$ is a constant.
Answer: a constant
Hence, the correct answer is option: 'D'.