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Homework Help: Average velocity vector

  1. Feb 23, 2015 #1
    1. The problem statement, all variables and given/known data

    A bicyclist makes a three quarter turn of radius 1.5 m at constant speed. During the turn, the only horizontal force acting on the bicyclist is a static friction force of constant magnitude 250 N. The total mass of the bicycle and rider is 60 kg.

    2. Relevant equations

    3. The attempt at a solution

    The path: http://www.mathgoodies.com/lessons/fractions/circles/circle_three_fourths_blue.gif

    The circle's radius = 1.5
    The bicyclist's speed: 2.5
    Acceleration at the point A: (-25/6,0)
    Velocity at the point A: (0, 2.5)
    Velocity at the point B (the final point of the journey): (2.5,0)
    And the average velocity (-0.53, -0.53)

    I have computed everything correctly except for average velocity and I have no idea how to get the right angle. I guessed there should be minus signs (-1.25, -1.25) but of course magnitudes are wrong.
    I realize that this is because of acceleration. You can't just reproduce strategies fit for linear motion, i.e. to divide the sum of final and initial velocity by two.

  2. jcsd
  3. Feb 23, 2015 #2


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    Staff: Mentor

    Look into the definition of average velocity. Hint: The velocities at the start and finish are irrelevant.
  4. Feb 23, 2015 #3
    I already tried this. I understand that this is the displacement over time. So displacement in this case is simply 1/4 out of the circumference. I divided it by the time I computed (constant velocity is given).

    Time = (3*((2*pi*1.5)/4))/2.5

    Average velocity (vector legnth) = (((2*pi*1.5)/4))/2.82743 = 0.833334
  5. Feb 23, 2015 #4


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    Staff: Mentor

    No. You seem to be trying to compute an average speed based upon the distance of the path. That's not average velocity.

    Forget the length of the path. You want to start by finding the displacement vector: the vector that describes the final position with respect to the starting location. Do it in vector form. If you take the origin as the center of the track, then the initial position is <r,0>. What's the final position in this form?
  6. Feb 23, 2015 #5
    Many thanks. :) I got it.

    initial position: (1.5,0)
    final position: (0, -1.5)

  7. Feb 23, 2015 #6


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    Staff: Mentor

    Very close. But your calculation of the displacement is not correct. You want to take the final position and subtract the initial position to find the net displacement:

    <displacement> = <final> - <initial>
  8. Feb 24, 2015 #7
    Oh, right. One magnitude would be positive, which is wrong.


    (-0.530517, -0.530517)
  9. Feb 24, 2015 #8
    this is decidedly a gap in my knowledge. I have to do more exercices.
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