You're an investigator for the National Transportation Safety Board, examining a subway accident in which a train going at \(80 \mathrm{km} / \mathrm{h}\) collided with a slower train traveling in the same direction at \(25 \mathrm{km} / \mathrm{h}\). Your job is to determine the relative speed of the collision, to help establish new crash standards. The faster train's "black box" shows that it began negatively accelerating at \(2.1 \mathrm{m} / \mathrm{s}^{2}\) when it was \(50 \mathrm{m}\) from the slower train, while the slower train continued at constant speed. What do you report?

Short Answer

Expert verified
The relative speed of the two trains at the point of their collision can be computed by following the steps above. This involves solving an equation of motion for time, calculating the final speed of the faster train using that time, and finding the relative speed by comparing this result with the constant speed of the slower train.

Step by step solution

01

Calculate Time

The initial velocity \(v_{0}\) of the faster train is \(80 \mathrm{km} / \mathrm{h}\) or \(22.22 \mathrm{m} / \mathrm{s}\). The acceleration \(a\) is \(-2.1 \mathrm{m} / \mathrm{s}^{2}\) (negative as the train is decelerating). The distance \(s\) it has to cover is 50 m. Implementing these values into the equation of motion \( s = v_{0}*t + \frac{1}{2}*a*t^{2}\), we can solve for \(t\).
02

Solve the Quadratic Equation

In solving for \(t\) we have a quadratic equation with \( t_{1,2} = \frac{-v_{0} \pm \sqrt{v_{0}^{2} - 2 * a * s}}{a} \( and we're interested in the positive root.
03

Find Final Velocity of Faster Train

After finding the time, we can then use it to find the final velocity of the faster train at the point of collision, using the formula \( v = v_{0} + a*t \).
04

Find Relative Speed

The speed of the slower train is \(25 \mathrm{km} / \mathrm{h}\) or \(6.94 \mathrm{m} / \mathrm{s}\). Subtract this speed from the final velocity of the faster train to get the relative speed.

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