Question

Displacement \( s \) of a particle at time \( t \) is expressed as \( s = 2t^3 - 9t \). Find the acceleration at the time when the velocity vanishes.

• A. \( 6 \)
• B. \( 6\sqrt{3} \)
• C. \( 6\sqrt{6} \)
• D. \( 3\sqrt{6} \)

Hint:

For kinematics problems, remember: - Velocity is the derivative of displacement: \( v = \frac{ds}{dt} \). - Acceleration is the derivative of velocity: \( a = \frac{dv}{dt} \). - To find acceleration when velocity is zero, first solve \( v = 0 \) for \( t \) and substitute in \( a(t) \).

Answer 1

The Correct Option is C

Step 1: Find the velocity function Velocity is the first derivative of displacement: \[ v = \frac{ds}{dt} = \frac{d}{dt} (2t^3 - 9t). \] Differentiating term by term: \[ v = 6t^2 - 9. \]
Step 2: Find the time when velocity vanishes Setting \( v = 0 \): \[ 6t^2 - 9 = 0. \] \[ 6t^2 = 9. \] \[ t^2 = \frac{9}{6} = \frac{3}{2}. \] \[ t = \sqrt{\frac{3}{2}} = \frac{\sqrt{6}}{2}. \]
Step 3: Find the acceleration function Acceleration is the derivative of velocity: \[ a = \frac{dv}{dt} = \frac{d}{dt} (6t^2 - 9). \] \[ a = 12t. \] Substituting \( t = \frac{\sqrt{6}}{2} \): \[ a = 12 \times \frac{\sqrt{6}}{2}. \] \[ a = 6\sqrt{6}. \] Thus, the correct answer is option (3) \( 6\sqrt{6} \).