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Homework Help: Linear Acceleration in a Circular Path

  1. Oct 28, 2007 #1
    1. The problem statement, all variables and given/known data

    A schoolbus travels around a circular path with acceleration a(t)=0.5t m/s/s with t in seconds.
    At some point it has a velocity of 8 m/s.
    What are the magnitudes of its velocity and acceleration when it has travelled a fourth of the circular track from the point at which it had v = 8 m/s?

    Radius of the track is 250m

    2. Relevant equations

    dv=adt
    ds=vdt

    3. The attempt at a solution

    Using the relation dv=adt,
    dv=0.5tdt

    I integrate both sides, but get confused as to where I should integrate from since the time the bus has a velocity of 8 m/s is not provided. After consideration, I choose:

    v initial is 0, v final is 8 and t initial is 0, t final is just t

    After integrating, I solve for t to find

    t=6.32 at v=10 and v(t)=0.25t^2

    then if a fourth of the circle is travelled, the distance travelled is a fourth of the circumference:

    (2*pi*250)/4=125*pi


    then ds=vdt=(0.03t^2)dt
    We integrate again to find distance travelled as a function of t
    integrate ds from 0 to 125*pi and vdt from 6.32 to t, yielding

    125*pi=0.083(t)^3-0.083(6.32)^3

    From this equation, t=17.08 seconds or it will take 17 seconds to travel a fourth of the way around from this point. From here, we can plug into the a(t) and v(t) equations to answer the question.

    I think this is right; however, the first time I solved the problem, I integrated the dv=adt expression from 10 to v on the dv side and 0 to t on the adt side to get v(t)=0.25*t^2+10
    I then integrated it again and found the distance as a function of time, solving the time it took to travel from the point at 10 m/s to a point a fourth of the way around the track. What is wrong with this approach?
     
  2. jcsd
  3. Oct 29, 2007 #2
    Your bounds for your first approach are strange. You are saying that

    [tex]\int_{10}^{v(t)} dv = \int_{10}^{t} a dt[/tex]

    but you don't know v(10).
     
  4. Oct 29, 2007 #3
    no, that's not correct.
    first, the acceleration integral is from starting point t=0.
    secondly, the velocity integral is going from 0 to 8 (or as I mistyped it, 10)
     
  5. Oct 29, 2007 #4
    also, it will take 17-6.32 seconds to travel that 200*pi
     
  6. Oct 30, 2007 #5
    I don't know if I confused you, or you're confusing me, but your second approach, the first half of your post is right.

    For your other method are you saying you wanted to do (?)

    [tex]\int_{v(0)}^{v(t)} dv = \int_0^t \frac{1}{2}t dt[/tex]

    that is what you did. If I take you literally (what was the typo again?), you are not considering that the bounds have to be the same on both sides of the integral

    [tex]\int_{v(t_1)}^{v(t_2)} dv = \int_{t_1}^{t_2} .5 t dt[/tex]

    You can't integrate velocity from 0 8, and time from t(2) to t(10). I don't know if this is what were trying to do. On the other hand, you could solve for time when velocity is 8 this way.

    [tex]\int_{0}^{8} dv = \int_{t(0)}^{t(8)} .5 t' dt'[/tex]

    If this doesn't clear things up, could you reformulate the other method you tried?
     
    Last edited: Oct 30, 2007
  7. Oct 30, 2007 #6
    I think that's what I did. The confusion is in part because of the 8-10 typo, but also because of the two methods, and the fact that I don't have the capability to make integrals easily. If you know a program that will do this, I'd appreciate it.

    The bounds and integrals that I used ended up being:

    0 to the final velocity, say 8....integrated dv
    equal to
    t=0 to t=x, x being the unknown time at which v has reached 8 for the integral of adt

    this gave me the time it initial took to go from rest to the point it was at v=8,
    at which point i could do the integral of

    d to d+125*pi the integral of ds
    equal to
    x to y(y being the time it took to travel 125*pi from the point v=8) the integral of vdt

    does this make sense? i'm doing my best to explain the method clearly, but it's hard wihout integral notation here.
     
  8. Oct 30, 2007 #7
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