Question

The position \( x \) (in meters) of a particle moving along a straight line is given by: \[ x = t^3 - 12t + 3 \] where \( t \) is time (in seconds). The acceleration of the particle when its velocity becomes \( 15 \, \text{ms}^{-1} \) is: \vspace{0.5cm}

• A. \( 15 \, \text{ms}^{-2} \)
• B. \( 24 \, \text{ms}^{-2} \)
• C. \( 18 \, \text{ms}^{-2} \)
• D. \( 12 \, \text{ms}^{-2} \) \vspace{0.5cm}

Hint:

To find acceleration from position-time equations, differentiate once to get velocity and again to get acceleration. Then substitute the given velocity to find the corresponding time.

Answer 1

The Correct Option is C

Step 1: Compute the Velocity Velocity is the first derivative of position \( x \) with respect to time: \[ v = \frac{dx}{dt} = \frac{d}{dt} (t^3 - 12t + 3) \] \[ v = 3t^2 - 12 \] We are given that velocity \( v = 15 \): \[ 3t^2 - 12 = 15 \] Solving for \( t \): \[ 3t^2 = 27 \] \[ t^2 = 9 \] \[ t = \pm3 \] Since time cannot be negative, we take \( t = 3 \). \vspace{0.5cm} Step 2: Compute the Acceleration Acceleration is the derivative of velocity: \[ a = \frac{dv}{dt} = \frac{d}{dt} (3t^2 - 12) \] \[ a = 6t \] Substituting \( t = 3 \): \[ a = 6(3) = 18 \, \text{ms}^{-2} \] \vspace{0.5cm}