Runner's power and displacement

  • Thread starter stinlin
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  • #1
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Homework Statement


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.


Homework Equations



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

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:

EDIT

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

So:

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:

Answers and Replies

  • #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.
 
  • #3

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

So:

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.


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.
 
  • #4
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Neither of those methods work - I tried them both previous to posting this. =(
 
  • #5
HallsofIvy
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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.
 
  • #6
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Woah, that gives the right answer...Haha - can you explain that one more time please?
 
  • #7
Dick
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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?
 
  • #8
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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?
 
  • #9
Dick
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How about a motor running at constant power and spinning up a flywheel? Stuff like that.
 
  • #10
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I can understand the verbal example, but what's the deal with integrating due to constant power development?
 
  • #11
Dick
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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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