Distance/time train problem

  • Thread starter Prio
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Homework Statement



A train is instructed to start at rest at station 1 and accelerate uniformly between points A and B, then coast with a uniform velocity between points B and C, and finally accelerate (slowing down) uniformly between points C and D until the train stops at station 2. The distances AB BC, and CD are all equal, and it takes 5.00 min to travel between the stations. Assume that the uniform accelerations have the same magnitude, even when they are opposite in direction.

Homework Equations



I'm not sure which equation to use here...

The Attempt at a Solution



change in x from AB = change in x from CD as does the value for a with opposite signs.
I set the total distance between points A and D to an arbitrary number...lets say 9, set acceleration to an arbitrary number, lets say 1 m/min, and tried to find a number for time that would satisfy the problem. So after one min, the train would reach 1 m velocity, then 2 m velocity after 2 min, and would have then traveled 3 m total which is 1/3 of the total distance. This would end up taking 2 min (AB) + 2 min (BD) + 1.5 min (CD) = 6.5 min to travel the distance of 9. I would need to find a value time where the train travels the same for BC as AB over a course of time that would make its distance work in 5 minutes total. Maybe I could use a common scale factor that take everything down proportionally. I think I made it more confusing with all that I just said...lol.

Surely there is a much simpler way to think it through?
 

Answers and Replies

  • #2
alphysicist
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Hi Prio,

Maybe I'm just missing it in your post, but what exactly are you looking for in this problem? The acceleration?
 
  • #3
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Well crap...you're right--I forgot to put that. So here...

You're trying to find amount of time between each leg of the trip, AB, BC, and CD, where presumably the times for AB and CD are the same.
 
  • #4
alphysicist
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Well crap...you're right--I forgot to put that. So here...

You're trying to find amount of time between each leg of the trip, AB, BC, and CD, where presumably the times for AB and CD are the same.
Okay, that's good. In this problem you'll need to write down (at least) three equations to solve for the two times. (Label all of the algebraic quantities that they don't give values for--let d be the distance between the points, and t1 be the time from A to B, and t2 be the time from B to C.)

I would start with BC. There's no acceleration, so what equation could you write down that relates d and the time t2? (When there is no acceleration there is only one equation that is helpful.)

Then write an equation for AB relating d and the time t1. Which of the five kinematic equations is best to use here? (You want to relate AB and BC algebraically, so get as many common variables in the equation as you can.)

Finally, you know the entire trip is 5 minutes, so write down an expression for T1 and T2 that shows this.

Do you get the right answer?
 
  • #5
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For BC I would use d = 1/2(initial v + final v )t2 but velocity doesn't change, so we could say
d = 1/2(2v)t = vt solving for t yields d/v--But should I be solving for d or t?

Then for AB I'm not sure which equation to use...I tried using d = 1/2(initial + final)t1 again, with initial as 0...so d=1/2(v)t1

Since the v is the same in both equation, it can be dropped...but then it yields d=t2 and d=1/2t1 which would be doubled so both distances get the same expression which is of course wrong. So then I don't really know which equation to use or if I've even done anything right...:\\\\\\

I also thought of using d=1/2 a (t1)2 and if I solve for t1 I get the sqrt of (2d/a) = t1

Put into 2t1 + t2 = 5, it makes some morass of inconclusive random numbers, so obviously I didn't do it right<_<
 
Last edited:
  • #6
alphysicist
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For BC I would use d = 1/2(initial v + final v )t2 but velocity doesn't change, so we could say
d = 1/2(2v)t = vt solving for t yields d/v--But should I be solving for d or t?
This is good--but keep it as

d= v t2


Then for AB I'm not sure which equation to use...I tried using d = 1/2(initial + final)t1 again, with initial as 0...so d=1/2(v)t1
Perfect so far--this is the second equation: d = (1/2) v t2

Since the v is the same in both equation, it can be dropped...but then it yields d=t2 and d=1/2t1
What did you do to drop v? The two equations you have written (d=t2 and d=1/2t1) are not correct (for one thing, they don't have the same units on each side of the equation).

You can leave the v in the equation, and eliminate d (or v) by solving one equation for d (or v) and plugging into the other. This will give one equation that has t1 and t2 in it that show how those two times are related to each other.

which would be doubled so both distances get the same expression which is of course wrong. So then I don't really know which equation to use or if I've even done anything right...:\\\\\\

I also thought of using d=1/2 a (t1)2 and if I solve for t1 I get the sqrt of (2d/a) = t1

Put into 2t1 + t2 = 5,
This last equation is the third equation you need: 2t1 + t2 = 5. If you then use the expression on how t1 and t2 are related (that you got by eliminating d or v), then you should be able to solve the problem.
 
  • #7
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thanks--
alright so if d = vt2 for BC and d = 1/2 vt1 for AB,

I'll plug (vt2) into the AB equation so that vt2 = 1/2 vt1 , which simplifies to t1 = 2 t2

Put into the last equation 2t1+t2=5, this means that 2 [2 t2] + t2 = 5

Now 5 t2 = 5 min

So apparently t2 = 1 min and t1 2 min

therefore...
t for AB & CD = 2 min, t for BC = 1 min

Hope that's right.

Thanks again for the help and quick replies. I may have to use this forum more often:P
 
  • #8
alphysicist
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That looks right to me.
 

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