Question

the distance s along z axis varies with time as a s = 2t 3 +3z 2 +8. find the time t after which acceleration becomes zero

Answer 1

The law of motion

$$s(t)=2t^3+3t^2+8$$

The acceleration

$$a(t)=\frac{d^2s(t)}{dt^2}=12t+6$$

$$12t+6=0$$

$$t=-0.5\:\rm{s}$$