Relative Motion in One Dimension — Question 5

Question 5

A direction reversal changes the closing rate

Point A begins at x=0 moving east at 12 m/s. Point B begins at x=300 m moving west at 8 m/s. After 5 s, B reverses instantaneously and then moves east at 8 m/s. Ignore the duration of the idealized reversal.

1. Find the separation just before B reverses.

2. Find the encounter time and position after the reversal.

3. Compare with the encounter time if B had continued west. Explain why a single closing speed cannot describe both stages.

Question 5 - Solution

A direction reversal changes the closing rate

(a) Close the gap during the first stage.

$$d(5)=300-(12+8)(5)=\boxed{200\,\mathrm m},\quad x_A=60,\quad x_B=260\quad\mathrm m$$ Initial closing speed is $12-(-8)=20$ m/s. In 5 s the gap falls by 100 m, leaving 200 m. The positions are $x_A=60$ m and $x_B=260$ m.

(b) Reset the relative-motion equation at the switch.

$$\tau=\frac{200}{12-8}=50\,\mathrm s,\quad T=5+50=55\,\mathrm s,\quad x=12(55)=660\,\mathrm m$$ Afterward, A gains at $12-8=4$ m/s, so closing the remaining 200 m takes 50 s. Total time is 55 s, and the encounter is at x=$12(55)=660$ m, also $260+8(50)$ m.

(c) Compare the uninterrupted approach.

$$T_{\mathrm{no\ reversal}}=300/(12+8)=15\,\mathrm s,\quad v_{A/B}:20\longrightarrow4\,\mathrm{m/s}$$ Without reversal they would meet after $300/20=15$ s. The reversal changes the signed relative velocity from 20 to 4 m/s. Applying either value to the entire 55 s history would incorrectly assume that B’s direction remained unchanged.