1. **Problem statement:** Two stations A and B are 800 m apart on a straight track. Train 1 passes A at $t=0$ moving towards B at 20 m/s. Train 2 passes B at $t=5$ moving towards A at 20 m/s, then after 10 s (at $t=15$) increases speed to 25 m/s. We need to: (a) Sketch displacement-time graph for both trains. (b) Find where and when they pass each other. (c) Determine which train reaches the opposite station first. 2. **Formulas and rules:** Displacement $s = s_0 + vt$ for constant speed. For piecewise speed, use $s = s_0 + v_1 t_1 + v_2 t_2$ where $t_1$ and $t_2$ are time intervals. 3. **Train 1 displacement:** Starts at A ($s=0$) at $t=0$, speed $v=20$ m/s. Displacement at time $t$ is: $$s_1(t) = 20t$$ 4. **Train 2 displacement:** Starts at B ($s=800$ m) at $t=5$, moving towards A (direction opposite to Train 1), so velocity is $-20$ m/s initially. For $5 \leq t < 15$: $$s_2(t) = 800 - 20(t - 5) = 800 - 20t + 100 = 900 - 20t$$ For $t \geq 15$: Train 2 increases speed to 25 m/s towards A. Displacement at $t=15$: $$s_2(15) = 900 - 20 \times 15 = 900 - 300 = 600$$ For $t \geq 15$: $$s_2(t) = 600 - 25(t - 15) = 600 - 25t + 375 = 975 - 25t$$ 5. **(a) Sketch displacement-time graph:** - Train 1: straight line from $(0,0)$ with slope 20. - Train 2: piecewise line starting at $(5,800)$ with slope -20 until $t=15$, then slope -25 after $t=15$. 6. **(b) Find when and where trains pass each other:** They meet when $s_1(t) = s_2(t)$. Check intervals: - For $5 \leq t < 15$: $$20t = 900 - 20t$$ $$20t + 20t = 900$$ $$40t = 900$$ $$t = \frac{900}{40} = 22.5$$ (not in $5 \leq t < 15$) - For $t \geq 15$: $$20t = 975 - 25t$$ $$20t + 25t = 975$$ $$45t = 975$$ $$t = \frac{975}{45} = 21.666...$$ (valid since $21.666... > 15$) 7. **Calculate meeting position:** $$s = s_1(21.666...) = 20 \times 21.666... = 433.33 \text{ m}$$ 8. **(c) Which train reaches opposite station first?** - Train 1 reaches B at $s=800$: $$20t = 800 \Rightarrow t = 40 \text{ s}$$ - Train 2 reaches A at $s=0$: For $t \geq 15$: $$0 = 975 - 25t$$ $$25t = 975$$ $$t = 39$$ Train 2 reaches A at $t=39$ s, Train 1 reaches B at $t=40$ s. **Answer:** - (a) Displacement-time graph as described. - (b) Trains pass each other at $t=21.67$ s approximately, at 433.33 m from A. - (c) Train 2 reaches opposite station first at 39 s.