1. Given position as a function of time

$x(t)=-3t^3+18t^2+16t$

We need the velocity when acceleration becomes zero.

2. Find velocity

Velocity is the time derivative of position:

$v(t)=\frac{dx}{dt}$

So,

$v(t)=\frac{d}{dt}(-3t^3+18t^2+16t)$ $v(t)=-9t^2+36t+16$

3. Find acceleration

Acceleration is the time derivative of velocity:

$a(t)=\frac{dv}{dt}$

Thus,

$a(t)=\frac{d}{dt}(-9t^2+36t+16)$ $a(t)=-18t+36$

4. Set acceleration equal to zero

When acceleration becomes zero,

$-18t+36=0$

$18t=36$

$t=2\ \text{s}$

5. Find velocity at $t=2$ s

Substitute $t=2$ into the velocity equation:

$v(2)=-9(2)^2+36(2)+16$

$v(2)=-9\cdot 4+72+16$

$v(2)=-36+72+16=52$

6. Final answer

The velocity when acceleration becomes zero is:

$\boxed{52\ \text{m/s}}$