As two trains move along a track, their conductors suddenly notice that they are headed towards each other. Figure 2-31 gives their velocitiesv as functions of time t as the conductors slow the trains. The figure’s vertical scaling is set by ${\mathit{v}}_{\mathbf{s}}\mathbf{=}\mathbf{40}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$. The slowing processes begin when the trains are 200 m apart. What is their separation when both trains have stopped?

Velocity of the first train $=40\text{m}/\text{s}$

Velocity of the second train $=30\text{m}/\text{s}$

The problem involves a simple mathematical operation to calculate the area of a triangle. According to the above graph, the displacement of each train is the area of a triangle. And the displacement is the integral of velocity. Thus, Using the formula for the area of a triangle, the displacements of both the trains and hence the total distance traveled by both trains can be calculated.

Formula:

Displacement = Area under the curve of velocity vs. time.

Area of triangle = $\frac{1}{2}\left(\mathrm{base}\right)\times \left(\mathrm{height}\right)$

For each train, displacement is the area in the graph. As each area is triangular,

Area of triangle = $\frac{1}{2}\left(\mathrm{base}\right)\times \left(\mathrm{height}\right)$

Therefore, displacement of first train is

$\frac{1}{2}\left(40\frac{\text{m}}{\text{s}}\right)\left(5\text{s}\right)=100\text{m}$

And, displacement of the other train is

$\frac{1}{2}\left(30\frac{\text{m}}{\text{s}}\right)\left(4\text{s}\right)=60\text{m}$

Initially, the trains were separated by 200 m and later both the trains traveled a distance

$100m+60m=160m$.

So, the gap between the trains becomes

$200\text{m}-160=40\text{m}$.

Therefore, the separation between the two trains is 40 m.