Philip P. answered • 01/28/16

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Let's break the problem up into three segments

- segment 1 is distance traveled while the train is accelerating at 1 m/s
^{2} - segment 2 is the distance traveled at constant speed - given as 2100 m
- segment 3 is the distance traveled while decelerating at 2 m/s
^{2}

In segment 1, the distance that the train moves while accelerating at 1 m/s

^{2}for 20 seconds is: distance

_{1}= (1/2)*a*t^{2}= (1/2)(1 m/s^{2})(20 s)^{2}=**200 m** The speed it attains is:

speed = a*t = (1 m/s

^{2})(20 s) = 20 m/sIn segment 2, we are given distance

_{2}= 2100 m.In segment 3, the train then decelerates at 2 m/s

^{2}from a speed of 20 m/s until it stops, speed = 0. So the train must decelerate by 20 m/s: a*t = 20 m/s

(2 m/s

^{2})*t = 20 m/s t = 10 sec

So the distance the train traverses while decelerating at 2 m/s

^{2}is: distance

_{3}= (1/2)*a*t^{2}= (1/2)(2 m/s^{2})(10 s)^{2}=**100 m**The total distance between the stations is the sum of the three distances:

distance between stations = distance

_{1}+ distance_{2}+ distance_{3} = 200 m + 2100 m + 100 m

=

**2400 m**