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Physics help - kinematics

  1. Sep 26, 2005 #1
    Problem 1.

    In a race a distance d long, A and B cross the finish line in a dead heat, both taking a time T. Accelerating uniformly, A takes a time ta and B a time tb to attain maximum speed, which they maintain for the rest of their race.

    a) what is the acceleration of each sprinter?
    b) what are their respective maximum speeds?

    For both questions, express your answer in terms of d, T, ta and tb.

    This is what I've done so far.. my logic is probably all wrong but any help is appreciated.

    1. First, I think that the equation of A and B will be very similar (only ta, tb and maximum speeds are different).

    2. I use the kinematic equations, and divide the race of A into two halves. The first half is when A is accelerating uniformly.

    Phase 1
    V2 = v1 + a(delta d)^2
    V2 = a(delta d)^2

    the distance travelled in phase 1 would be (v2 + v1)/2 x ta
    d1 = (v2/2)ta

    Phase 2
    d2 = v2(T - ta)

    Can someone tell me if my logic up to this point is correct?
     
  2. jcsd
  3. Sep 26, 2005 #2
    you're not using the correct formula it is

    [tex] v_{2}^2 = v_{1}^2 + 2ad [/tex]
    good idea doing it in phases i did it the same way. So for the first leg of A you have uniform acceleration where u use the above formula
    For the second leg of A you have constant velocity with time where time is
    T - Ta (do you see why?) the total time - the time taken to reach is the time A will be at cosntant velocity. Now Add the distances from both phases. And isolate for aceleration. Do the same for B.
     
  4. Sep 26, 2005 #3
    Sorry I am stuck again.

    After I added the two phases togehter, I am left with the following equation

    dtotal = (v2)^2 / 2a + v2 ( T - Ta)

    how would i solve for a?
     
  5. Sep 27, 2005 #4

    Fermat

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    I assume what you have is,

    [tex]d_{total} = \frac{v_2^2}{2a} + v_2 ( T - T_a)[/tex]

    where a is the acceleration during the first phase and v2 is the final velocity, with v1 = 0.

    you will have gotten v2 = aTa - just substitute.
     
  6. Sep 27, 2005 #5
    Got the help I needed, thanks!
     
    Last edited: Sep 28, 2005
  7. Sep 28, 2005 #6

    Fermat

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    Yes, you did it right - those answers are correct.

    btw, your answer for the acceln might look neater if the t1 was taken out as a factor :smile:

    [tex]
    a = \frac{2d}{t_{1}(2T - t_{1})}[/tex]
     
  8. Sep 28, 2005 #7
    thank you very much! I really appreciate your help =)
     
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