Velocity and Acceleration — Question 9

Question 9 – Solution

Strategy Minimize squared distance; it has the same minimizer as distance and avoids a square root.

Relative position $$\vec d(t)=\vec r_1-\vec r_2=\left\langle 2t-4,2t-1,-2t\right\rangle.$$ Thus $$D^2(t)=(2t-4)^2+(2t-1)^2+(-2t)^2=12t^2-20t+17.$$

Minimum The vertex occurs at $t=20/24=5/6$. Then $$D^2\left(\frac 56\right)=\frac{26}{3}, \qquad \boxed{D_{\min}=\sqrt{\frac{26}{3}}}.$$ Because the quadratic opens upward and $5/6\ge 0$, this is the required global minimum.