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Runner's power and displacement

  1. Feb 19, 2007 #1
    1. The problem statement, all variables and given/known data
    A 60-kg runner increases her speed from 2 m/s to 4.3 m/s in 5 s. Assuming she develops constant power during this time interval and neglecting air resistance, determine (a) the power developed, (b) the distance traveled.

    2. Relevant equations

    P = W/t
    T = kinetic energy = 1/2mv^2

    3. The attempt at a solution

    The power was easy:

    T1 + U1->2 = T2

    U1->2 = T2 - T1 = 1/2*m(v2^2 - v1^2) = 434.7 J

    P = U/t = 434.7/5 = 86.9 W

    Ok, but the displacement, I'm not sure:

    P = W/t = Fd, and that's all I have. I don't know F. I know P, W,t, but without F or d, I'm slightly screwed. Can I have some help? If it helps, I DO know the answer, but I'd like to see a solution/arrive at an answer before looking at it. :rolleyes:


    I just tried Impulse-Momentum and STILL didn't get the right answer:

    mv1 + Imp 1->2 = mv2

    Imp 1->2 = F(t2-t1) and t2-t1 = 5s


    F = (m(v2-v1))/(t2-t1) = 27.6

    P = W/t = Fd/t -----> Pt/F = d

    Using that, I'm getting the wrong answer of 15.74 m.
    Last edited: Feb 19, 2007
  2. jcsd
  3. Feb 19, 2007 #2
    You can use the kinematics equations to solve for the distance.
    [tex]s=ut+0.5at^2[/tex], you have t, you can find out a by (v-u)/t.
  4. Feb 19, 2007 #3

    The forces here arent impulsive. You could equate the change in momentum to average force, divide by m to get average acceleration and then use the equation above to get the distance.
  5. Feb 19, 2007 #4
    Neither of those methods work - I tried them both previous to posting this. =(
  6. Feb 19, 2007 #5


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    "Assuming she develops constant power", then at any time t, 0< t< 5, ((1/2)mv2- (1/2)m(22)/t= 30(v2- 4)/t= 86.9 so
    v2= 2.9t+ 4. [itex]v= dx/dt= \sqrt{2.9t+ 4}[/itex]. Integrate that to find the distance run in those 5 seconds.
  7. Feb 19, 2007 #6
    Woah, that gives the right answer...Haha - can you explain that one more time please?
  8. Feb 19, 2007 #7


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    He's using the 'constant power' condition to say that the energy of the runner is increasing at a constant rate. In other words, that kinetic energy is a linear function of time. Meaning in turn that v^2 is a linear function of time. Using information in the problem he finds this function, solves for v and integrates it. Does that help?
  9. Feb 19, 2007 #8
    I guess it makes perfect sense, I'm just confused still about the idea of constant power development and it's implications. Can you give another example where this would apply by chance?
  10. Feb 19, 2007 #9


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    How about a motor running at constant power and spinning up a flywheel? Stuff like that.
  11. Feb 19, 2007 #10
    I can understand the verbal example, but what's the deal with integrating due to constant power development?
  12. Feb 19, 2007 #11


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    There's no special concept of 'constant power development'. The point was just to find the time derivative of something and then integrate it to get the change in the something.
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