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: Understanding the Concept:
Velocity is \(v = \dfrac{ds}{dt}\) and acceleration is \(a = \dfrac{dv}{dt}\). We are told \(s\) is proportional to \(v^2\).

Step 2: Key Formula or Approach:
Write \(s = c\,v^2\) for a constant \(c\), then differentiate both sides with respect to time.

Step 3: Detailed Explanation:
\[ \frac{ds}{dt} = 2c\,v\,\frac{dv}{dt} \]
Replace \(\dfrac{ds}{dt}\) by \(v\) and \(\dfrac{dv}{dt}\) by \(a\):
\[ v = 2c\,v\,a \]
For a moving particle \(v \neq 0\), so we may cancel \(v\):
\[ a = \frac{1}{2c} \]
This is a constant, since \(c\) is a constant. So the acceleration does not change with time or with displacement.
This matches uniformly accelerated motion, where \(v^2 = 2as\). Options (A), (B) and (C) say that \(a\) depends on \(s\), which would not be the case here.

Final Answer:
The acceleration is a constant, option (D). \[ \boxed{\text{a constant (D)}} \]