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Homework Help: Circular Motion of two bikes

  1. Jan 24, 2004 #1
    Two cyclists A and B, are traveling counterclockwise around a circular track at a constant speed of 8 ft/sec at the instant shown. If the speed of A is increased at aA = SA ft/sec^2, where SA is in ft, determine the distance measured counterclockwise along the track from B to A between the cyclists when time is = 1 sec. What is the magnitude of the acceleration of each cyclist at the instant?

    To find the length of an arc, you use the equation arc=theta*radius, but how do you encorporate the time into this? I don't know how far cyclist A moves.

    To find the magnitude of acceleration you can use the sqrt of a(normal)^2 + a(tangential)^2.

    aB =1.28 ft/sec^2

    because a(tangential)=0 (constant velocity) and a(normal)=64/50

    I don't know what aA is equal to.

    Any suggestions?

    I posted a picture too.
     

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    Last edited: Jan 24, 2004
  2. jcsd
  3. Jan 25, 2004 #2

    Doc Al

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    Staff: Mentor

    First find the distance between A & B at time t=0 (the instant shown). Then find out how far each moves in the next second. If it wasn't for that fact that A is accelerating, they would move the same distance, thus maintaining the same separation. But A gains some distance over B: ΔX = 1/2at2.

    Also, A gains some speed: ΔV = at.
     
  4. Jan 25, 2004 #3
    at time t=0 the distance between them is 104.72ft

    B moves 8ft in the next second because it is constant

    but i still don't get how far A moves.

    and how do I use deltaX and deltaV
     
    Last edited: Jan 25, 2004
  5. Jan 25, 2004 #4

    Doc Al

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    Staff: Mentor

    You need to understand the basic formulas for uniform accelerated motion. One key relation is [itex]d = v_0t + \frac{1}{2}at^2[/itex], which describes the distance traveled in time t. ([itex]v_0[/itex] is the initial speed.) Another useful formula gives the speed after time t: [itex]v = v_0 + at[/itex]. You will need both of these to understand how "A" moves.

    B is just moving at a constant speed (tangential acceleration = 0). I believe you understand that, but note that the above equations apply if you set a=0. (Please try this!)

    Of course, the above only applies to the tangential motion. To find the full acceleration, you must add the centripetal acceleration.

    Note: Δ just means "change"; ΔX means change in x.
     
  6. Jan 25, 2004 #5
    Thanks!

    I got the final answers.

    Thank-you.
     
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