Question

Question attached as screenshot below: hard calculus, please only answer if you feel you are capable.

Answer 1

The position vector is given by

$S(t)=t^2+3t+1$

The velocity will be the derivative of the position vector.

$\begin{gathered} v(t)=2t+3 \\ v(4)=11\text{mps} \end{gathered}$

The acceleration will be

$a(t)=\frac{dv}{dx}=2mps^2$

So velocity is 11 m/s, acceleration is 2 m/s^2