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Race Acceleration problem

  1. Sep 22, 2004 #1
    Ok this is another problem I am having trouble with.

    In a 100 m race, Maggie and Judy cross the finish line in a dead heat, both taking 10.2 s. Accelerating uniformly, Maggie took 2.10 s and Judy 3.10 s to attain maximum speed, which they maintained for the rest of the race. What was Maggie's acceleration?

    Ok so I know that

    Si (initial distance)= 0 m
    Sf (final distance)= 100m
    Delta T = 10.2s (10-0)

    Vi( Initial velocity of maggy and Judy) = 0 m/s

    I think Judy's time is there just to distract us, because every time I try to use both set of data and plug them into a kinematic equations for motion with constant acceleration, I reach a dead end.

    So I tried a different method

    100m - Sf = 1/2 * a * (2.10s) ^ 2

    100m - Sf = 2.205A

    A = (100m-Sf)/2.205

    And Then I tried to plug that back into the SF equatoin but that didn't help.


    Can anyone point me to the right direction?
     
  2. jcsd
  3. Sep 22, 2004 #2
    In a 100 m race, Maggie and Judy cross the finish line in a dead heat, both taking 10.2 s. Accelerating uniformly, Maggie took 2.10 s and Judy 3.10 s to attain maximum speed, which they maintained for the rest of the race. What was Maggie's acceleration?

    Since the acceleration isn't constant throughout the race, you can't simply use 1/2a*t^2.

    So maggie's position would be something like this:

    1/2a*(2.1)^2 + vf(10.2-2.1) = 100m

    You can find vf by saying vf = a*t where t is how much time she spent accelerating. I hope I didn't give away too much!
     
  4. Sep 22, 2004 #3
    So you have

    1. Time taken by both to reach the finish line = 10.2 seconds (since this is a dead heat)
    2. Distance covered by both in 10.2 seconds = 100m
    3. Suppose Vm and Vj are the speeds of Maggie and Judy at any time t, then Vm is maximum at 2.10s and Vj is maximum for 3.10s.

    Now draw a graph of Vm versus t and Vj versus t. Find the maximum speed of both Maggie and Judy (given the initial speed = 0 m/s). After t = Tmax--different for Maggie and Judy--the speeds are constant and equal to their respective values at 2.10s and 3.10s. Does that give you some idea how to draw the graph? (If not, the v-t graph has a linearly increasing section followed by a line parallel to the time axis).

    Okay up to here.

    Thats probably because your analysis isn't quite right. And you get equations which aren't too friendly either.

    I don't see what you are doing. Sf has a different meaning here does it? Otherwise, from what you wrote above, 100m = Sf and so each of these equations makes no sense.

    Well I think you should graphically do this first, because the solution isn't very difficult and can be done in more than one way. You need two equations for the two accelerations, given the above conditions. Try and rework this after making the graph and follow up with your solution.

    Cheers
    Vivek
     
  5. Sep 22, 2004 #4

    Gokul43201

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    Staff Emeritus
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    Gold Member

    I thought you were saying Sf = 100 m ?

    Anyway, like you said, forget about judy.

    Split the sprint into two sections : (1) the acceleration , (2) the cruise

    What do you know about section (1) ? The time taken, t = 2.1 s. Call the distance s, the acceleration a, and the final velocity v. What equations can you write relating these quantities.

    Now for section (2), the distance run is 100 - s, and is run at constant velocity v(accel = 0). You alsko know the time taken for this stretch. Can you not relate 100 - s to v and the time ?

    You now have 3 equations in 3 unknowns. Solve them.
     
    Last edited: Sep 22, 2004
  6. Sep 22, 2004 #5
    Oh sorry with these equations

    100m - Sf = 1/2 * a * (2.10s) ^ 2
    100m - Sf = 2.205A
    A = (100m-Sf)/2.205

    I meant Sf = The distance travelled in 2.10 seconds
    But I just realized that even if I got the Acceleration from that, It would for after the 2.10 seconds. And I know that anyway. Since after 2.10 seconds they maintained the maximum speed the velocity must be increasing constantly and so the A must be 0.

    Ok so that went out the window.

    Now the equations that I can relate this three (distance s, the acceleration a, and the final velocity v) are

    S = (Vf-vi) * delta t (since vi = 0, it's Vf * delta T )
    S = 1/2 a delta t ^ 2
    A = Vf * delta T

    Now from the info on section 2, now I know that

    100 - S = 1/2 a * delta t ^ 2 Or

    100 - (vf * delta T) = 1/2 a * delta t ^ 2 or

    100 - (Vf (10.2- 2.21) = 1/2 * a * 4.41

    100 - (vf (7.99) = 2.205A

    100 = 2.205A + Vf(7.99)

    100 = 2.205A + A * (2.10) (7.99)

    100 = 18.984A

    A = 5.27 m/s/s

    Yay, Thank you guys for the help.
     
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