Two trains, each having a speed of 24 km/h, are headed at each other on the same straight track. A bird that can fly 48 km/h flies off the front of one train when they are 48 km apart and heads directly for the other train ?

Sep 8, 2015

The bird travels 32 km until it meets the other train.

Explanation:

The speed of the train carrying the bird is irrelevant.

Once the bird leaves the train, it can fly at only 48 km/h.

The bird and the train will be travelling for the same time until they meet.

Since $s = v t$, $t = \frac{s}{v}$

Let $1$ represent the train and $2$ represent the bird.

Since ${t}_{1} = {t}_{2}$, and

${s}_{1} / {v}_{1} = {s}_{2} / {v}_{2}$,

${s}_{2} / {s}_{1} = {v}_{2} / {v}_{1} = \left(48 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{km/h"))))/(24 color(red)(cancel(color(black)("km/h}}}}\right) = 2$

(1) ${s}_{1} = {s}_{2} / 2$, and

(2) ${s}_{2} + {s}_{1} = \text{48 km}$

Substitute Equation (1) into Equation (2):

${s}_{2} + {s}_{2} / 2 = \text{48 km}$

$\frac{3 {s}_{2}}{2} = \text{48 km}$

s_2 = 2/3 × "48 km"

${s}_{2} = \text{32 km}$