1. **Problem statement:** Jeff and Will run back and forth between two parallel lines A and B, 120 m apart. Jeff starts at line A running towards B at 4 m/s. Will starts at line B running towards A at 6 m/s. We need to find the times and positions where they pass each other for the first three times. 2. **Key idea:** Both run back and forth between lines A and B, reversing direction instantly at each line. 3. **Period of each runner:** - Jeff's one round trip distance: $2 \times 120 = 240$ m - Jeff's speed: 4 m/s - Jeff's period: $T_J = \frac{240}{4} = 60$ s - Will's one round trip distance: $240$ m - Will's speed: 6 m/s - Will's period: $T_W = \frac{240}{6} = 40$ s 4. **Position functions:** Define $x=0$ at line A and $x=120$ at line B. - Jeff starts at $x=0$ going towards B: For $t$ in $[0,60)$, Jeff's position is: $$x_J(t) = \begin{cases} 4t & 0 \leq t < 30 \\ 120 - 4(t-30) & 30 \leq t < 60 \end{cases}$$ This pattern repeats every 60 s. - Will starts at $x=120$ going towards A: For $t$ in $[0,40)$, Will's position is: $$x_W(t) = \begin{cases} 120 - 6t & 0 \leq t < 20 \\ 6(t-20) & 20 \leq t < 40 \end{cases}$$ This pattern repeats every 40 s. 5. **Find first meeting time $t_1$:** They meet when $x_J(t) = x_W(t)$. For $t$ in $[0,20)$ (both going towards each other): $$4t = 120 - 6t \implies 10t = 120 \implies t = 12 \text{ s}$$ Check $t=12$ is in $[0,20)$ for Will and $[0,30)$ for Jeff, valid. 6. **Position at first meeting:** $$x = 4 \times 12 = 48 \text{ m from line A}$$ Distance from line B: $120 - 48 = 72$ m. 7. **Find second meeting time $t_2$:** Next, consider intervals where their directions might differ. Between $t=20$ and $t=30$: - Jeff: still going towards B, $x_J(t) = 4t$ - Will: running back towards B, $x_W(t) = 6(t-20)$ Set equal: $$4t = 6(t-20) \implies 4t = 6t - 120 \implies 2t = 120 \implies t = 60$$ But $t=60$ is outside this interval. Check next intervals: Between $t=30$ and $t=40$: - Jeff: running back towards A, $x_J(t) = 120 - 4(t-30)$ - Will: running back towards B, $x_W(t) = 6(t-20)$ Set equal: $$120 - 4(t-30) = 6(t-20)$$ $$120 - 4t + 120 = 6t - 120$$ $$240 - 4t = 6t - 120$$ $$240 + 120 = 6t + 4t$$ $$360 = 10t \implies t = 36$$ Check $t=36$ in $[30,40)$ valid. 8. **Position at second meeting:** $$x = 120 - 4(36-30) = 120 - 24 = 96 \text{ m from line A}$$ Distance from line B: $120 - 96 = 24$ m. 9. **Find third meeting time $t_3$:** Next intervals: Between $t=40$ and $t=60$: - Jeff: running back towards A, $x_J(t) = 120 - 4(t-30)$ - Will: running towards A again (new cycle), $x_W(t) = 120 - 6(t-40)$ Set equal: $$120 - 4(t-30) = 120 - 6(t-40)$$ $$-4t + 120 = -6t + 240$$ $$-4t + 6t = 240 - 120$$ $$2t = 120 \implies t = 60$$ Check $t=60$ in $[40,60)$ for Will? Will's cycle restarts at 40, so $t=60$ is in $[40,80)$ for Will's second cycle. 10. **Position at third meeting:** $$x = 120 - 4(60-30) = 120 - 120 = 0 \text{ m from line A}$$ Distance from line A: 0 m. **Final answers:** - First meeting: $t_1 = 12$ s, 48 m from line A (72 m from B) - Second meeting: $t_2 = 36$ s, 96 m from line A (24 m from B) - Third meeting: $t_3 = 60$ s, 0 m from line A (at line A)