Question

a particle moves along a straight line such that its displacement s at time t is given by s=t3 - 6t2+3t+4 meter find teh velocity when the acceleration is zero

Answer 1

The displacement is given as:
$$\begin{gathered} s=t^3-6t^2+3t+4 \\ \text{The velocity is given as:} \\ v=\frac{ds}{dt}=\frac{d}{dt}(t^3-6t^2+3t+4)=3t^2-12t+3 \\ \text{The acceleration is given as:} \\ a=\frac{dv}{dt}=\frac{d}{dt}(3t^2-12t+3)=6t-12 \\ \text{When the acceleration is zero,} \\ a=0 \\ \Rightarrow 6t-12=0 \\ \Rightarrow t=12/6=2\text{s} \\ \text{Therefore, velocity at}t=2s, \\ v=3t^2-12t+3=3\times 2^2-12\times 2+3=12-24+3=-9\text{m/s} \end{gathered}$$