Question

Two trains start at the same time from Delhi and Jalandhar, distance of 400 km, travelling, one at the rate of 48 km/hr and the other at 72 km/hr. Where will they meet and in what time from starting.

Answer 1

Answer on Question #43667 – Physics - Mechanics | Kinematics | Dynamics

Two trains start at the same time from Delhi and Jalandhar, distance of $400\mathrm{km}$, travelling, one at the rate of $48\mathrm{km/hr}$ and the other at $72\mathrm{km/hr}$. Where will they meet and in what time from starting.

Solution:

$S = 400\mathrm{km}$ – distance between two trains;

$V_{1} = 48\frac{\mathrm{km}}{\mathrm{hr}}$ – velocity of the first train;

$V_{2} = 72\frac{\mathrm{km}}{\mathrm{hr}}$ – velocity of the first train;

Let trains are moving towards and the distance travelled by the train with slower speed be $x$. Accordingly the distance travelled by the faster train will be $y = S - x$. Let the time taken by the trains to travel the distance be $t$

According to question:

$$\text{first (slower)train: } t = \frac{x}{V_1}$$ $$\text{second (faster)train: } t = \frac{y}{V_2} = \frac{S - x}{V_2}$$ $$(1) = (2):$$ $$\frac{x}{V_1} = \frac{S - x}{V_2}$$ $$V_2x = SV_1 - V_1x$$ $$x(V_1 + V_2) = SV_1$$ $$x = \frac{SV_1}{V_1 + V_2} = \frac{400\mathrm{km} \cdot 48\frac{\mathrm{km}}{\mathrm{hr}}}{48\frac{\mathrm{km}}{\mathrm{hr}} + 72\frac{\mathrm{km}}{\mathrm{hr}}} = 160\mathrm{km}$$ $$y = S - x = 400\mathrm{km} - 160\mathrm{km} = 240\mathrm{km}$$ $$t = \frac{x}{V_1} = \frac{160\mathrm{km}}{48\frac{\mathrm{km}}{\mathrm{hr}}} = 3 \text{ hours } 20 \text{ minutes}$$

Answer: trains will meet $160\mathrm{km}$ from station where slower train has been started to move; they will meet after 3 hours 20 minutes.

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