Relative Motion in Two Dimensions — Question 5

Question 5

Intercept a moving target by aiming ahead

A target begins at $(30,0)$ m and moves north at 4 m/s. A pursuer starts at the origin at the same time and may choose a constant velocity of magnitude 5 m/s. Both move in one horizontal plane, with no acceleration.

1. Use a common arrival time T to find the required pursuer velocity components.

2. Find the interception time, location and heading.

3. Explain why aiming directly at the target’s initial position would miss it.

Question 5 - Solution

Intercept a moving target by aiming ahead

(a) Require equality of complete position vectors.

$$\vec v_PT=\vec r_{T,0}+\vec v_TT=(30,4T)\ \Rightarrow\ \vec v_P=(30/T,4)\quad\text{(SI)}$$ At T the target is at $(30,4T)$ m, so the pursuer needs $\vec v_P=(30/T,4)$ m/s, with T in seconds. Its north component must match the target, leaving the east component to close the original offset.

(b) Impose the fixed speed.

$$(30/T)^2+4^2=5^2\ \Rightarrow\ \boxed{T=10\,\mathrm s,\quad\vec r_{\mathrm{hit}}=(30,40)\,\mathrm m},\quad\theta=53.130^\circ$$ $(30/T)^2+4^2=5^2$ gives $T=10$ s for positive time. Thus $\vec v_P=(3,4)$ m/s, heading $53.130^\circ$ north of east, and interception is at $(30,40)$ m.

(c) Check the moving destination.

$$T_{\mathrm{due\ east}}=30/5=6\,\mathrm s,\quad\vec r_T(6)-\vec r_P(6)=(0,24)\,\mathrm m\ne\vec0$$ A due-east 5 m/s launch would reach $(30,0)$ m after 6 s, when the target is already at $(30,24)$ m. It reaches the target’s old location, not its simultaneous location. Aiming ahead compensates for the target displacement during the travel time.