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Expert: - 7/7/2009


Question
A train travels from A to B,a distance of 20000m,taking 1000s.The journey has 3 stages.In the first stage the train starts from rest at A & accelerates uniformly until its speed is Vms^-1.In the second stage the train travels at constant speed Vms^-1 for 600s.During the third stage of the journey the train decelerates uniformly,coming to rest at B.Given that the acceleration of the train during the first stage of the journey is 0.15ms^-2,find the distance travelled by the train during the third stage of the journey.Thanks

Answer
Note that I don't know if there is an exact answer.

I am sending you what I have worked out so far.

At least, I'm sending you what looks important out of a half dozen pages, most of which lead me astray.

To start off with, I came up with a list of variables.

 Let a = 0.15m/s².
 It occurs only at the start and not in the middle.

 Let v be the velocity in the middle.
 It is of little use at the start or end.

 Let (t1) = starting time, (t2) = middle time,
 and (t3) = ending time.

 Let (d1), (d2), and (d3) be the distances of the respective times.

Note that v = a(t1) and (d1) = a*(t1)²/2 = 0.075(t1)².

Also note that the distance travelled in the middle is
(d2) = v*(t2) = (0.15*(t1))(600) = 90*(t1).

The distance left is (d3) = 20,000 - (d1) - (d2)
= 20,000 - 0.075*(t1)² - 90(t1).

The distance left is also (d3) = v(t3) - a(t3)²/2.

The time left is (t3) = 1000 - (t1) - (t2)
= 1000 - (t1) - 600 = 400 - (t1).

Using the equation just given for (d3), it can be seen
(d3) = (0.15*(t1))(400 - (t1)) - 0.15(400 - (t1))²/2.

Note that (d3) must be less than 0.15*(a3)²/2.

That's about all I can think of right now,
but right back after studying this answer
if you've got more question.

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