Question

If the velocity of a particle moving along a straight line with uniform acceleration, is given by \( V = \sqrt{196 - 16X} \, \text{m/s} \), then its acceleration is

• A. \( 8 \, \text{m/s}^2 \)
• B. \( 14 \, \text{m/s}^2 \)
• C. \( -8 \, \text{m/s}^2 \)
• D. \( -16 \, \text{m/s}^2 \)

Hint:

When acceleration is required, use the chain rule to relate velocity to displacement and time.

Answer 1

The Correct Option is C

We are given the velocity of the particle as a function of displacement: \[ V = \sqrt{196 - 16X} \] To find the acceleration, we first recall that acceleration \( a \) is the derivative of velocity with respect to time: \[ a = \frac{dV}{dt} \] Using the chain rule, we can express this as: \[ a = \frac{dV}{dX} \cdot \frac{dX}{dt} = \frac{dV}{dX} \cdot V \] Now, differentiate \( V = \sqrt{196 - 16X} \) with respect to \( X \): \[ \frac{dV}{dX} = \frac{-8}{\sqrt{196 - 16X}} \] Therefore, the acceleration is: \[ a = \frac{-8}{\sqrt{196 - 16X}} \cdot \sqrt{196 - 16X} = -8 \, \text{m/s}^2 \] Thus, the acceleration is \( -8 \, \text{m/s}^2 \). \[ \boxed{-8 \, \text{m/s}^2} \]