A particle is moving in one dimension (along the x-axis) under the action of a variable force. Its position \( x \) (in meters) varies with time \( t \) (in seconds) according to the equation \( x = -3t^3 + 18t^2 + 16t + 16 \). The velocity of the particle when its acceleration becomes zero is _________ \( \text{m/s} \).

Explanation:
1. Finding Velocity: Velocity \( v \) is the first derivative of position with respect to time (\( v = \frac{dx}{dt} \)):
\( v = \frac{d}{dt}(-3t^3 + 18t^2 + 16t + 16) \)
\( v = -9t^2 + 36t + 16 \)

2. Finding Acceleration: Acceleration \( a \) is the first derivative of velocity with respect to time (\( a = \frac{dv}{dt} \)):
\( a = \frac{d}{dt}(-9t^2 + 36t + 16) \)
\( a = -18t + 36 \)

3. Condition for Zero Acceleration: Setting the acceleration equation to zero allows us to determine the specific time \( t \):
\( -18t + 36 = 0 \)
\( 18t = 36 \implies t = 2\text{ s} \)

4. Calculating Velocity at \( t = 2\text{ s} \):
Substituting \( t = 2 \) back into the velocity equation yields:
\( v(2) = -9(2)^2 + 36(2) + 16 \)
\( v(2) = -9(4) + 72 + 16 \)
\( v(2) = -36 + 72 + 16 = 52\text{ m/s} \)

Hence, Option B is the correct choice.