Light-rail passenger trains that provide transportation within and between citie

Question

Light-rail passenger trains that provide transportation within and between cities are capable of modest accelerations. The magnitude of the maximum acceleration is typically 1.3 , but the driver will usually maintain a constant acceleration that is less than the maximum. A train travels through a congested part of town at 5.0 . Once free of this area, it speeds up to 14 in 8.0 . At the edge of town, the driver again accelerates, with the same acceleration, for another 16 to reach a higher cruising speed. What is the final speed?

Explanation / Answer

This time, you need to know that acceleration × time = change in velocity. So, for example, imagine a stopped car that accelerates at 10 m/s² for 5 seconds. Let's concentrate on the acceleration of 10 m/s² because that's usually the confusing part—and also the key to understanding the physics! It means that the car's velocity increases by 10 m/s during every second that the car is accelerating. So after 1 second, it will be going 10 m/s. After 2 seconds, it be going 2s × 10 m/s², or 20 m/s. And after 5 seconds, it will be going 50 m/s. Ok, now let's apply that to the light-rail problem. The train left the congested area at 4.0 m/s accelerated to 4 m/s. So it sped up by 14 – 4 = 10 m/s, and this happened after accelerating at a constant rate for 8 s. That means its velocity increased by 10 m/s in 8 s, or 10/8 = 5/4 = 1¼ m/s². So far, so good? Next, when it left the town, it accelerated at the same rate for another 16s. In other words, it started at 14m/s and then sped up by 5/4 m/s every second, for 16 s. So its final speed is 14 m/s + (5/4 m/s² × 16 s)

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