Velocity and Acceleration — Question 7

Question 7

A particle has velocity $$\mathbf v(t)=\left\langle t-2,2t+1,2\right\rangle,\qquad t\ge 0.$$ Tasks

1. Find when its speed is smallest.

2. Find the minimum speed.

3. Show that velocity is perpendicular to acceleration at that instant and explain why.

Question 7 – Solution

Strategy. Minimize squared speed, which has the same minimizers as speed and avoids a square root.

Step 1: Squared speed $$\begin{align*} q(t)=\|\mathbf v(t)\|^2 &=(t-2)^2+(2t+1)^2+4\\ &=5t^2+9. \end{align*}$$ On $t\ge 0$, this is minimized at $\boxed{t=0}$.

Step 2: Minimum speed $$\mathbf v(0)=\left\langle -2,1,2\right\rangle, \qquad \boxed{\|\mathbf v(0)\|=3}.$$

Step 3: Orthogonality Acceleration is $$\mathbf a(t)=\mathbf v'(t)=\left\langle 1,2,0\right\rangle.$$ At $t=0$, $$\mathbf v(0)\cdot\mathbf a(0)=-2+2+0=0.$$ Also $q'(t)=2\mathbf v\cdot\mathbf a=10t$, so the vanishing dot product is exactly the zero derivative of squared speed. Here the minimum occurs at the endpoint, and the right-hand derivative is nonnegative immediately afterward.