How Can Uniform Acceleration and Velocity Affect Train Travel Times?

[Prio]

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, let's 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?

 

[alphysicist]

Hi Prio,

Maybe I'm just missing it in your post, but what exactly are you looking for in this problem? The acceleration?

 

[Prio]

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.

 

[alphysicist]

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?

 

[Prio]

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<_<

 

[alphysicist]

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.

 

[Prio]

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

 

[alphysicist]

That looks right to me.