1. **Problem statement:** Two stations A and B are 800 m apart on a straight track. - Train 1 passes station A at $t=0$ moving towards B at 20 m/s. - Train 2 passes station B at $t=5$ moving towards A at 20 m/s. - After 10 seconds from $t=5$ (i.e., at $t=15$), Train 2 increases speed to 25 m/s. We need to find: (a) Displacement-time graph for both trains. (b) Time and position where trains pass each other. (c) Which train reaches the opposite station first. 2. **Formulas and rules:** - Displacement $s = s_0 + vt$ for constant speed. - For piecewise speed, use separate equations for each time interval. - Trains move towards each other, so their positions add up to 800 m when they meet. 3. **Train 1 position:** Starts at A ($s=0$) at $t=0$, speed $v_1=20$ m/s. $$s_1(t) = 20t$$ 4. **Train 2 position:** Starts at B ($s=800$) at $t=5$, moving towards A (direction opposite to Train 1), so displacement decreases. - For $5 \leq t < 15$ (first 10 seconds after $t=5$): $$s_2(t) = 800 - 20(t-5) = 800 - 20t + 100 = 900 - 20t$$ - For $t \geq 15$ (speed increases to 25 m/s): At $t=15$, position: $$s_2(15) = 900 - 20 \times 15 = 900 - 300 = 600$$ For $t \geq 15$: $$s_2(t) = 600 - 25(t-15) = 600 - 25t + 375 = 975 - 25t$$ 5. **(b) Find meeting time and position:** They meet when $s_1(t) = s_2(t)$. - For $0 \leq t < 15$, use $s_2(t) = 900 - 20t$: $$20t = 900 - 20t$$ $$20t + 20t = 900$$ $$40t = 900$$ $$t = \frac{900}{40} = 22.5$$ Since $22.5 > 15$, meeting is not in this interval. - For $t \geq 15$, use $s_2(t) = 975 - 25t$: $$20t = 975 - 25t$$ $$20t + 25t = 975$$ $$45t = 975$$ $$t = \frac{975}{45} = 21.666\ldots \approx 21.67$$ This is valid since $21.67 > 15$. Position at meeting: $$s = 20 \times 21.67 = 433.33 \text{ meters from A}$$ 6. **(c) Which train reaches opposite station first?** - Train 1 reaches B at $s=800$: $$20t = 800 \Rightarrow t = 40 \text{ seconds}$$ - Train 2 reaches A at $s=0$: For $t \geq 15$: $$0 = 975 - 25t$$ $$25t = 975$$ $$t = 39$$ Train 2 reaches A at $t=39$ seconds, earlier than Train 1 reaching B at $t=40$ seconds. **Final answers:** - (a) Displacement-time graph: Train 1 is a straight line from $(0,0)$ to $(40,800)$. Train 2 is piecewise linear: from $t=5$ to $15$, line from $(5,800)$ to $(15,600)$, then from $15$ onwards line with steeper slope to $0$ at $t=39$. - (b) Trains meet at $t \approx 21.67$ seconds, $433.33$ meters from A. - (c) Train 2 reaches station A first at $t=39$ seconds.