1. **Problem statement:** Two stations A and B are 800 m apart on a straight track. Train 1 passes station A at time $t=0$ moving towards B at 20 m/s. Train 2 passes station B at time $t=5$ moving towards A at 20 m/s, then after 10 seconds (at $t=15$) increases speed to 25 m/s in the same direction. We need to sketch displacement-time graphs for both trains on the same diagram. 2. **Formulas and rules:** Displacement $s$ as a function of time $t$ for constant speed $v$ is: $$s = s_0 + v(t - t_0)$$ where $s_0$ is initial position at time $t_0$. 3. **Train 1 motion:** - Starts at $s=0$ m (station A) at $t=0$. - Moves towards B at 20 m/s. So displacement for train 1 is: $$s_1 = 0 + 20t = 20t$$ valid for all $t \geq 0$. 4. **Train 2 motion:** - Starts at $s=800$ m (station B) at $t=5$. - Moves towards A (opposite direction) at 20 m/s from $t=5$ to $t=15$. - Then speed increases to 25 m/s towards A for $t > 15$. For $5 \leq t \leq 15$: $$s_2 = 800 - 20(t - 5) = 800 - 20t + 100 = 900 - 20t$$ For $t > 15$: At $t=15$, position is: $$s_2(15) = 900 - 20 \times 15 = 900 - 300 = 600$$ Then for $t > 15$: $$s_2 = 600 - 25(t - 15) = 600 - 25t + 375 = 975 - 25t$$ 5. **Summary of displacement functions:** Train 1: $$s_1 = 20t, \quad t \geq 0$$ Train 2: $$s_2 = \begin{cases} 900 - 20t, & 5 \leq t \leq 15 \\ 975 - 25t, & t > 15 \end{cases}$$ 6. **Interpretation for graph:** - Train 1 starts at 0 m at $t=0$ and moves linearly upwards with slope 20. - Train 2 starts at 800 m at $t=5$, moves linearly downwards with slope -20 until $t=15$, then slope changes to -25. This completes the displacement-time model for both trains. **Final answer:** Train 1 displacement: $$s_1 = 20t$$ Train 2 displacement: $$s_2 = \begin{cases} 900 - 20t, & 5 \leq t \leq 15 \\ 975 - 25t, & t > 15 \end{cases}$$