Relative Motion in One Dimension — Question 3

Question 3

A delayed pursuit changes the starting separation

A runner A starts at x=0 and runs east at 7 m/s. Runner B starts 120 m ahead and runs east at 5 m/s. Treat both as points moving at constant speeds.

1. If both start at the same time, find the catch time and position.

2. If A waits 10 s while B continues, find the catch time measured from A’s departure and from B’s departure.

3. Explain why dividing the original 120 m gap by 7 m/s gives neither catch time.

Question 3 - Solution

A delayed pursuit changes the starting separation

(a) Use the rate at which the gap closes.

$$120+(5-7)t=0\ \Rightarrow\ \boxed{t=60\,\mathrm s,\quad x=7(60)=420\,\mathrm m}$$ The relative velocity is $7-5=2$ m/s. With simultaneous starts the catch time is $120/2=60$ s, at x=$7(60)=420$ m. B has moved during the pursuit.

(b) Update the gap before resetting the clock.

$$d_{\mathrm{delayed}}=120+5(10)=170\,\mathrm m,\quad\tau=170/2=85\,\mathrm s,\quad t_B=95\,\mathrm s$$ During A’s wait, B travels 50 m, so the gap at A’s departure is 170 m. Catch occurs 85 s after A starts, or 95 s after B starts. The position is $7(85)=595$ m, also $120+5(95)$ m.

(c) Keep target motion in the model.

$$\frac{120}{7}\ne\frac{120}{7-5},\qquad\frac{d}{dt}(x_B-x_A)=5-7=-2\,\mathrm{m/s}$$ $120/7$ would be the time for A to reach B’s initial position if that position were the target. It ignores both B’s ongoing motion and, in the delayed case, the extra 50 m head start. The denominator for closing a moving gap is the relative speed.