Question

The relation between velocity \( V \) (in ms\(^{-1}\)) and the displacement \( x \) (in meters) of a particle in motion is \( 2V = \sqrt{37 + 32x} \). 
The acceleration of the particle is 
 

• A. 32 ms\(^{-2}\)
• B. 8 ms\(^{-2}\)
• C. 16 ms\(^{-2}\)
• D. 4 ms\(^{-2}\)

Hint:

To find acceleration from a velocity-displacement relation, differentiate the equation with respect to time and use the definitions of velocity and acceleration.

Answer 1

The Correct Option is D

Given the relation: \[ 2V = \sqrt{37 + 32x} \] Square both sides to eliminate the square root: \[ (2V)^2 = 37 + 32x \] \[ 4V^2 = 37 + 32x \] Differentiate both sides with respect to time \( t \): \[ \frac{d}{dt}(4V^2) = \frac{d}{dt}(37 + 32x) \] \[ 8V \cdot \frac{dV}{dt} = 32 \cdot \frac{dx}{dt} \] We know that \( \frac{dx}{dt} = V \) (velocity) and \( \frac{dV}{dt} = a \) (acceleration). Substituting these values: \[ 8V \cdot a = 32V \] Divide both sides by \( 8V \): \[ a = \frac{32V}{8V} = 4 \, {ms}^{-2} \] Thus, the acceleration of the particle is 4 ms\(^{-2}\). 
Final Answer:  4 ms\(^{-2}\)