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) \( 6\sqrt{6} \).

Answer 2

To find the acceleration at the time when the velocity vanishes, we should first calculate the velocity and then find the time when it becomes zero. To begin, we find the velocity by taking the derivative of the displacement function \( s(t) = 2t^3 - 9t \). The velocity \( v(t) \) is given by:

\( v(t) = \frac{ds}{dt} = \frac{d}{dt}(2t^3 - 9t) = 6t^2 - 9 \)

To find when the velocity vanishes, set \( v(t) = 0 \):

\( 6t^2 - 9 = 0 \)

Solve for \( t \):

\( 6t^2 = 9 \)

\( t^2 = \frac{9}{6} = \frac{3}{2} \)

\( t = \pm\sqrt{\frac{3}{2}} = \pm\frac{\sqrt{6}}{2} \)

Since time cannot be negative, consider \( t = \frac{\sqrt{6}}{2} \).

Next, we find the acceleration by taking the derivative of the velocity:

\( a(t) = \frac{dv}{dt} = \frac{d}{dt}(6t^2 - 9) = 12t \)

Now, substitute \( t = \frac{\sqrt{6}}{2} \) into the acceleration function:

\( a\left(\frac{\sqrt{6}}{2}\right) = 12\left(\frac{\sqrt{6}}{2}\right) \)

\( = 6\sqrt{6} \)