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Need help with one dimensional kinematics problem

  • #1
Hello,
I've been working on the following problem for quite some time, with no success:
A commuter train travels between two downtown stations. Because the stations are only 1.15 km apart, the train never reaches its maximum possible cruising speed. The engineer minimizes the time t between the two stations by accelerating at a rate a1 = 0.105 m/s^2 for a time t1 and then by braking with acceleration a2 = -0.585 m/s^2 for a time t2. Calculate the minimum time of travel t.​

I've approached the problem in a few different ways; i've been attempting to state the supplied information in such a way that I can treat it as a max/min optimization problem and differentiate from there. I've been setting up the information using the equation:
xfinal = xinitial + velinitial(t) + 1/2(a)t^2

for example where for the first interval t1:
xfinal = 1150m
xinitial = 0
vinitial = 0
accel = 0.105m/s^2
time = the unknown

I'm running into the conceptual problem of how to deal with the two intervals. Where the time as I've laid it out is referring to just one of the two intervals, where I need to be dealing with both, i.e. the total time. I hope someone can give me an idea of how I'm misconceiving this problem. I feel like there needs to be another equation relating the two time intervals, but I can't figure out how I need to state the variables and how to integrate both equations.
I hope I've shown enough here to incline someone to give me some guidance, because I've really run up against a brick wall on this one, and I've spent a lot of time with it.
Thank you for your time!
 

Answers and Replies

  • #2
mukundpa
Homework Helper
524
3
Vmax = 0.105*t1 = 0.585*t2 ------------------------(1)

1150 = 0.5*0.105*t1^2 + 0.5*0.585*t2^2 -----------(2)

Think of these two equations. can yau solve them?
 
  • #3
Thanks for your reply, but I'm not sure how to proceed, honestly. Though I do see the relation that you've made between t1 and t2 in terms of Vmax, I really don't understand why you've done this. How is knowing the max velocity pertinent?
I can work the equations, solving equation (1) for t1, subbing into equation 2, then using that result in the original equation to find the other interval, then adding the result. But that doesn't give me a correct answer, unfortunately. And ultimately, I don't understand why this method should work, which is most important to me. If someone could further clarify the methodology behind a solution, I'd really appreciate it. Thanks.
 
  • #4
Fermat
Homework Helper
872
1
can you tell us what the correct answer is ?
 
  • #5
70
1
I think it's best to set up two equations: one for the total distance traveled, and one for the final velocity of the train. You'll get:

[tex]x_{f}=\frac{1}{2}a_{1}t_{1}^{2} + v_{f,1} t_{2} - \frac{1}{2}a_{2}t_{2}^{2}[/tex]

where the subscript 'f' denotes 'final' and the subscripts '1' and '2' denote the part of the trip where the train is accelerating and the part of the trip where the train is decelerating, respectively. [tex]a_{1}[/tex] and [tex]a_{2}[/tex] are both positive in this case. Also, [tex]v_{f,1}[/tex] can be substituted with [tex]a_{1}t_{1}[/tex], so the only two unknowns left in this equation are [tex]t_{1}[/tex] and [tex]t_{2}[/tex]. The second equation, for the final velocity is:

[tex]v_{f} = a_{1}t_{1} - a_{2}t_{2}[/tex].

Of course, [tex]v_{f} = 0[/tex]. Again, in this equation, both [tex]a_{1}[/tex] and [tex]a_{2}[/tex] are taken positive. Now you have a system of two equations with two unknowns, [tex]t_{1}[/tex] and [tex]t_{2}[/tex]. This system can be solved without further problems.
 
  • #6
Actually, no. It's a problem in an online course, you enter your final numerical answer and it either accepts it as correct, or doesn't. My final answer using the method above, was 68.23 s, and I've tried interpreting it in a few different ways, arriving at different answers, but none seem to jive.
Does this method seem correct? If so, could you explain it a a little bit?
Thank you.
 
  • #7
Brinx said:
I think it's best to set up two equations: one for the total distance traveled, and one for the final velocity of the train. You'll get:

[tex]x_{f}=\frac{1}{2}a_{1}t_{1}^{2} + v_{f,1} t_{2} - \frac{1}{2}a_{2}t_{2}^{2}[/tex]

where the subscript 'f' denotes 'final' and the subscripts '1' and '2' denote the part of the trip where the train is accelerating and the part of the trip where the train is decelerating, respectively. [tex]a_{1}[/tex] and [tex]a_{2}[/tex] are both positive in this case. Also, [tex]v_{f,1}[/tex] can be substituted with [tex]a_{1}t_{1}[/tex], so the only two unknowns left in this equation are [tex]t_{1}[/tex] and [tex]t_{2}[/tex]. The second equation, for the final velocity is:

[tex]v_{f} = a_{1}t_{1} - a_{2}t_{2}[/tex].

Of course, [tex]v_{f} = 0[/tex]. Again, in this equation, both [tex]a_{1}[/tex] and [tex]a_{2}[/tex] are taken positive. Now you have a system of two equations with two unknowns, [tex]t_{1}[/tex] and [tex]t_{2}[/tex]. This system can be solved without further problems.
Thank you, thank you, thank you, Brinx! Your reply clears things up immensly, I'll work it through and see if I can get the correct answer.
 

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