Velocity and Acceleration — Question 1

Question 1 – Solution

Strategy Differentiate componentwise and use the sign of $\vec v\cdot\vec a$.

Kinematics $$\vec v=\left\langle 2t-2,3t^2-3,2\right\rangle,\qquad \vec a=\left\langle 2,6t,0\right\rangle.$$ At $t=2$, $\vec v=\left\langle 2,9,2\right\rangle$, $\vec a=\left\langle 2,12,0\right\rangle$, and $$\boxed{\text{speed}=\|\vec v\|=\sqrt{89}}.$$

Speeding test $\vec v\cdot\vec a=4+108=112>0$, so speed is increasing. The particle is at $t=2$.