A particle is moving in one dimension (along x axis) under the action of a variable force. It's initial position was 16 m right of origin. The variation of its position (x) with time (t) is given as `x = –3t^3 + 18t^2 + 16t`, where x is in m and t is in s. The velocity of the particle when its acceleration becomes zero is _________ m/s.

Answer

To solve the problem, we need to find the velocity of a particle when its acceleration becomes zero. We are given the position function of the particle as:

\[ x(t) = -3t^3 + 18t^2 + 16t \]

### Step 1: Find the velocity function
The velocity \( v(t) \) is the first derivative of the position function \( x(t) \) with respect to time \( t \).

\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(-3t^3 + 18t^2 + 16t) \]

Using the power rule for differentiation:

\[ v(t) = -9t^2 + 36t + 16 \]

### Step 2: Find the acceleration function
The acceleration \( a(t) \) is the derivative of the velocity function \( v(t) \) with respect to time \( t \).

\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(-9t^2 + 36t + 16) \]

Again, applying the power rule:

\[ a(t) = -18t + 36 \]

### Step 3: Set the acceleration to zero
To find the time \( t \) when the acceleration is zero, we set the acceleration function equal to zero:

\[ -18t + 36 = 0 \]

Solving for \( t \):

\[ 18t = 36 \\ t = 2 \, \text{s} \]

### Step 4: Find the velocity at \( t = 2 \)
Now that we have the time when the acceleration is zero, we substitute \( t = 2 \) back into the velocity function to find the velocity at that moment.

\[ v(2) = -9(2^2) + 36(2) + 16 \]

Calculating this step by step:

\[ v(2) = -9(4) + 72 + 16 \\ v(2) = -36 + 72 + 16 \\ v(2) = 36 + 16 \\ v(2) = 52 \, \text{m/s} \]

### Final Answer
The velocity of the particle when its acceleration becomes zero is:

\[ \boxed{52 \, \text{m/s}} \]