Two trains, each having a speed of 30 km/h, are headed at each other on the same straight track. A bird that can fly 60 km/hflies off the front of one train when they are 60 kmapart and heads directly for the other train. On reaching the other train, the bird flies directly back to the first train and so forth. (We have no idea why a bird would behave in this way). What is the total distance the bird travels before the trains collide?

The speed of each train,$s_{\mathrm{train}}=30\mathrm{km}/\mathrm{h}$.

The speed of the bird,$s_{\mathrm{bird}}=60\mathrm{km}/\mathrm{h}$.

The initial distance between the trains, X=60 Km .

Speed may be defined as the time rate of change in distance. The time to cover the distance before the collision is used to find the distance traveled by the bird.

The expression for the speed is given as follows:

$\mathrm{Speed}=\frac{\mathrm{Distance}}{\mathrm{Time}}$ (i)

Here, both the trains travel at the same speed so that they will collide midway.

So, the distance covered by the train before the collision is calculated as follows:

$\begin{array}{c}x\text{'}=\frac{x}{2}\\ =\frac{60\mathrm{km}}{2}\\ =30\mathrm{km}\end{array}$

From equation (i), the time to cover this distance is as follows:

$\begin{array}{c}t=\frac{\mathrm{Distance}}{s_{\mathrm{train}}}\\ =\frac{30\mathrm{km}}{30\mathrm{km}/\mathrm{h}}\\ =1\mathrm{h}\end{array}$

Again, using equation (i), the distance traveled is calculated as follows:

$\begin{array}{c}\mathrm{Distance}=s_{\mathrm{bird}}\times t\\ =60\mathrm{km}/\mathrm{h}\times 1\mathrm{h}\\ =60\mathrm{km}\end{array}$

Hence, the total distance the bird travels before the trains collide is 60 Km.