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Solution anyone?

  1. Jul 6, 2007 #1
    A train has a length of 92m and starts from rest with a constant acceleration at time t=0s. At this instant, a car just reaches the end of the train. The car is moving with a constant velocity. At a time t=14s, the car just reaches the front of the train. Ultimately, however, the train pulls ahead of the car, and at time t=28s, the car is again at the rear of the train. Find the magnitudes of (a) the car's velocity and (b) the train's acceleration.

    I cannot find the required information to solve this.

    Train: Car:
    Initial v=0 V=constant
    Initial t=0 X=?
    acceleration= constant T=?
    final velocity@t=28s

  2. jcsd
  3. Jul 6, 2007 #2


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    Try to put it into equational form.

    - What do you know about the location of the car (say, [tex]s_\mathrm{car}(t)[/tex])?
    - What do you know about the location of the train (say, [tex]s_\mathrm{train}(t)[/tex])?
    - Which equations can you make? You should have as many equations as unknowns (I got two), so that you can solve for the unknowns ([itex]v_0[/itex] and [itex]a_0[/itex]). Try to expres is in formulae.

    Then tell us, where do you get stuck now?
    Last edited: Jul 6, 2007
  4. Jul 6, 2007 #3


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    Welcome to the PF, jaysun. As CompuChip is implying, you must show us your own work in order for us to provide tutorial help. We do not give out answers here on the PF.

    Are you familiar with the kinematic equations of motion? Those are what you use to solve these types of questions....

  5. Jul 6, 2007 #4
    sorry guys...and this isn't my homework. I am trying to teach myself general physics. So I do not have someone who knows to help. I have an old book from college back in 2001 that I am using. Honestly I have written down all the knowns and unknowns, but I do not know exactly where to start. I understand the problem, but cannot see what I am missing to get started. All that I know is that the displacement of the train and the car are equal at t=0, at that point the car is at a constant velocity(unknown) and the train starts to accelerate (unknown) from zero constantly. Then at t=14s the car travels the length of the train(92m) plus(+) the displacement (x) of the train at t= 14s. Then the train's velocity finally overtakes the car's velocity and the car is at the rear of the train again @ t=28s....

    I know the equations of kinematics, but I cannot find 3 of the 5 components needed to solve. I know that the car is a constant velocity formula ( v= x/t) and the train is and equation of kinematics.

    a= v/t
    a= v-v0/t-t0
    x= (v+v0/2)(t)
    x= 1/2at²v0t
    v= at+v0
    t= v-v0/a
  6. Jul 6, 2007 #5


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    No worries, jaysun. But since you are self-learning the material, let's keep it in the Homework Help forums.

    On your question, are you familiar with graphs? That might help to give you some intuition as to how to solve this. I'd start by drawing an distance versus time plot (time t on the horizontal axis, and distance d on the vertical). At time t=0 (the whole d axis) put the head of the train on d=92m, and the tail at d=0. Put the car also at d=0 at time t=0 (the place where the two axes cross).

    Now, the car is moving with some constant velocity, so draw a straight line up and to the right from the origin. The slope of this line (actually the rise divided by the run) is delta_d/delta_t, or the change in position with respect to time, which is the velocity, right? Since the velocity is constant, this line is straight. Don't worry that you don't know the actual velocity number yet -- we're just making a plot to help figure out the problem.

    Now, the train moves differently. Since the head and tail are connected rigidly, we can just make parallel plot lines for each. They will stay the same delta_d distance apart (92m) the whole time. So plot two parallel parabolas, the tail position parabola starts at the origin, and goes up and to the right with a shape of d=Ct^2, where C is some coefficient. Draw a similar parabola up an to the right from the head of the train, starting at t=0 and d=92m.

    Now, do you see how the car is even with the tail of the train at first, then at some time (we are told 14s), it is even with the head (and maybe passes it for a bit? I don't know), but the train keeps accelerating to higher speeds than the car, and the straight car line gets crossed one last time by the parabola line for the tail of the train?

    Okay, now that you have the graph, does that help you to see how to set up the equations? The 4th equation that you list above is the right one to start with, but what is Vo for the train if it starts at rest?

    I think you'll end up with a couple simultaneous equations, and solving those should get you the answers. I'll check back in a bit to see how you're doing.
  7. Jul 7, 2007 #6
    Sorry...even after plotting the graph I still have the same information that I knew before...

    @t=0 the car and tail end of the train(v0=0) are x=0
    and the head is at x=92m
    @t=14s the car is at the head of the train, with the tail 92m behind
    @t=28s the car is at the end of the train.

    I can use x= 1/2at²v0t and just plug in components but I don't understand why.

    By using this equation and don't know (a) we are looking for the train's acceleratoin, but the train didn't travel 92m @ t=14s... At t=14s the car travel 92m+ whatever the train traveled...? :confused:
  8. Jul 7, 2007 #7


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    You're getting closer.
    At time t, the car has travelled C(t) = v t, with v the constant speed of the car. The tail of the train, at that moment is at T(t) = (1/2) a t2, with a the constant accelleration of the train. Now what do you get if you plug in what you know at, say, t = 28? That is, what can you deduce from C(28) = T(28)?
  9. Jul 7, 2007 #8
    I don't completely understand your general variables... i know that x=vt, so I guess x=C(t) and x=1/2at² so T(t)=x.

    between t=14s and t=28s the train accelerates an additional 92m, which is +6.57m/s. If is increased it's speed by 6.57m/s then the acceleration should be (6.57m/s)/14s= 0.47m/s²? That would make displament of the car @14s 46.06m?

    So the a of the train is .47m/s²?
    and v of the car is 3.29m/s?
    Last edited: Jul 7, 2007
  10. Jul 7, 2007 #9


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    There are two different x's.
    There is the distance from the car to the origin, which is given by v t at a time t, and there is the distance from the back of the train to the origin, which is given by (1/2) a t2. Of course in general, these two distances are different, therefore you cannot call them both x. So I used C(t) to denote the distance travelled by the car at a time t, and T(t) for the distance from the back of the train to the origin at time t. I use these variables to express the positions of the car and train, for example, if I want to say that at the start of the measurement, they are at the same place, I'd write [itex]C(t = 0) = T(t = 0)[/tex]. If I want to express your first requirement, it would be [itex]C(t = 14) = T(t = 14) + 92[/itex]: the position of the car at time 14s is equal to the position of the (back of the) train at 14s + the length of the train.

    Now the trick is, that it is possible to these distances in terms of your unknowns, the velocity of the car and the acceleration of the train.

    Also note, that you cannot say
    Acceleration says how fast the velocity changes. So you can say, the train travels another so-and-so many meters, or it's velocity increases by so-and-so-much meters per second.
    Last edited: Jul 7, 2007
  11. Jul 7, 2007 #10
    ok, now I understand what you were saying. C(t=0) = T(t=0), C(t=14) = T(t=14)+92, and C(t=0) = T(t=28). :blushing: I was confused by the first "C is some coefficient" :uhh: The train travels 92m faster than the car in the last 14s. So now I have 3 variables x, t, and v0. So I use x=1/2at²+v0t

    0s 14s 28s

    a= 2(x-v0t)/t²
    a= 2(92)/14²

    So the train at 28s will be at a displacement of 368.5m

    Since the car and train are equal distances at t=28s, the velocity of the car is v=368.5m/28s=13.16m/s
  12. Jul 7, 2007 #11
    LOL, the answer was right under my nose the entire time. Thanks Compuchip and Berkeman
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