Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Need help with two 1d motion problems

  1. Sep 9, 2008 #1
    Problem 1
    A commuter train travels between two downtown stations. Because the stations are only 1.00 km apart, the train never reaches its maximum possible cruising speed. During rush hour the engineer minimizes the travel interval Δt between the two stations by accelerating for a time interval Δt1 at a1 = 0.100 m/s2 and then immediately braking with acceleration a2 = -0.440 m/s2 for a time interval Δt2. Find the minimum time interval of travel Δt and the time interval Δt1.

    So far what I have is that
    [tex] a_{1} t_{1} = a_{2} t_{2} [/tex]
    [tex] x_{f} = 1000 = \frac{1}{2} a_{1} t_{1}^2 + a_{1} t_{1} + a_{2} t_{2}^2 [/tex]
    I solved for T1 and substituted it in the second equation and got T1=159.59s and T2=36.27... no need to say those are wrong. I guess I am at a loss, I have been working this problem and the next for about 3 hours and I cannot find the correct solution.

    Problem 2
    Kathy Kool buys a sports car that can accelerate at the rate of 5.40 m/s2. She decides to test the car by racing with another speedster, Stan Speedy. Both start from rest, but experienced Stan leaves the starting line 1.00 s before Kathy. Stan moves with a constant acceleration of 3.70 m/s2, and Kathy maintains an acceleration of 5.40 m/s2.
    (a) Find the time it takes Kathy to overtake Stan.
    (b) Find the distance she travels before she catches him.
    (c) Find the speeds of both cars at the instant she overtakes him.

    Now on this problem I thought I understood it... I was wrong.
    I took the integral of the acceleration to get a V equation
    [tex] \frac{1}{2} 3.7t_{1}^2 = \frac{1}{2} 5.4t_{2}^2 [/tex]
    and I assumed [tex] t_{1} = t_{2} - 1 [/tex]
    and substituted in again to get T=5.81s ... which was wrong, and unfortunately I have used all my attempts for option A of this question, so to proceed I really need help on it.

    Thank you very much in advance if you take the time to help me.
  2. jcsd
  3. Sep 10, 2008 #2


    User Avatar
    Homework Helper

    Hi Prometheos,

    This equation is not correct. The second term on the right does not have the correct units. Check over how you derived that equation, and if you can't find the error please post how you derived it.

    This last relationship between [itex]t_1[/itex] and [itex]t_2[/itex] does not look right to me. The time [itex]t_1[/itex] is Stan's time, and the time [itex]t_2[/itex] is Kathy's time. But Stan has been in motion longer than Kathy. Do you see how this last equation needs to written?
  4. Sep 10, 2008 #3
    Looking back at my notes I actually have
    [tex] \frac{1}{2} 5.4t_{1}^2 = \frac{1}{2} 3.7t_{2}^2 [/tex]
    So that is my fault on converting it to latex. And, I think that is what you were suggesting in fixing my time relationship, it should be right now. But, the answer is still wrong =/
  5. Sep 10, 2008 #4


    User Avatar
    Homework Helper

    Okay, so [itex]t_1[/itex] is Kathy's time and and [itex]t_2[/itex] is Stan's time. And so now the relationship you used


    is true. But when you say you solved for the time and found 5.81 seconds, you called it T. What is T: is it [itex]t_1[/itex] or [itex]t_2[/itex]? And which one is the problem asking for? I think if you answer those questions you'll get the right answer.
  6. Sep 10, 2008 #5
    Ok, I had a response typed and in doing so I got it, finally my brain clicked. T was the total time or t2 (stan's), the question wanted the time it took kathy to overtake stan => t1=5.81 -1 = 4.81. I guess after hours of staring at one problem you start to miss the obvious.

    On Problem one I see now what you are saying, dimensional analysis gives me m/s when I want m. The trouble is, I don't know what equations to set up to answer this one. I thought that
    [tex] x_{f} = 1000 = \frac{1}{2} a_{1} t_{1}^2 + \frac{1}{2} a_{2} t_{2}^2 + x_{2o} [/tex]
    would work where [tex] x_{2o} [/tex] is the initial position when the train starts slowing down, or in other words the final x position after the train stops speeding up. But the problem I run into is I can't seem to find away to make the equations work with the information provided.
    Last edited: Sep 10, 2008
  7. Sep 10, 2008 #6


    User Avatar
    Homework Helper

    Actually [tex]x_{2o}=\frac{1}{2} a_{1} t_{1}^2[/tex], so the first term on the right side takes care of the [itex]x_{2o}[/itex].

    That's because [tex]\frac{1}{2} a_{1} t_{1}^2[/tex] is the displement from the first part of the motion, so it tells you where the train is at the beginning of the second part of the motion.

    (I think that you are also missing a numerical factor in the other term, and there is still a third term on the right that needs to be there.)

    Why don't you post how you derived your equation from the kinematic equations?
  8. Sep 10, 2008 #7
    as far as where I derived them from I just took the integral of velocity.
    I know the accelerations and I know they are over two different time intervals. And, I know the total distance is 1000m so from vf = vo + at I got 1000 = .5at^2
    But, that is the limit of my knowledge on this problem. It is just part of a review for my test tomorrow but I can't get it. I am missing the "hinge" to the problem that my teacher speaks of. I can't come up with a frame of reference on it, if I knew the max velocity achieved I could get time. If I knew the distance of the individual parts I could get time. But, I don't know those so I am lost. Probably 4 hours on this problem alone are lost as well.
  9. Sep 10, 2008 #8


    User Avatar
    Homework Helper

    I don't believe that's right. Once you integrate [itex]v = v0 + a t[/itex], with [itex]a[/itex] constant, you should get one of the standard kinematic formulas:

    x = x_0 + v_0 t +\frac{1}{2} a t^2

    You need to apply these two equation separately to the two parts of the motion. Get expressions for the "final" velocity and position of the speeding up part of the motion (in terms of [itex]a_1[/itex] and [itex]t_1[/itex]; these expressions will be the "initial" velocity and position for the two equations of the slowing down motion. What do you get?
  10. Sep 10, 2008 #9
    Ok finally got it, thanks to you.
    I set a1t1 = a2t2 => t2 = a1t1 / a2
    [tex] x_{f1} = 0 + 0 + .5(.1)t_{1}^2 [/tex]
    [tex] 1000 = .5(.1)t_{1}^2 + .1t_{1}t_{2} - .5(.44)t_{2}^2 [/tex]
    So the term I was missing was the [tex] .1t_{1}t_{2} [/tex]
    Substitution blah blah t1 = 127.72 t2 = 28.99

    Thanks for the help hopefully I get above a C on this test =/
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook