Power output of a person during a run

AI Thread Summary
To calculate the average power output of a person running 100m in 9.69 seconds with constant acceleration, the mass of the runner is essential but not provided in the problem. The discussion emphasizes the need to first determine acceleration using the formula d = 1/2 at^2, assuming initial velocity is zero. Once acceleration is known, the final velocity can be calculated, leading to the kinetic energy gained, which represents the work done. The power can then be calculated using P = W/t, but without the mass, the problem remains unsolvable. The conversation highlights the importance of mass in determining work and power in physics problems.
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Homework Statement


a person ran 100m in 9.69 seconds. assuming the acceleration was constant, what was his average power output during that run in watts?


Homework Equations


P=Fd/t or P= change in energy/t


The Attempt at a Solution


i don't actually know where to start... i know i need to find the force he used, but i only know how to do that using friction, which i don't have i also don't have his mass. it would also be really helpful if someone could tell me how i need use the information "assuming the acceleration was constant"
 
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You must first calculate acceleration, using the equation

d= vit + 1/2 at^2

We will assume vi = 0. We know d and t.

Then we can find the final velocity of the runner using the formula

vf = vi + at

Now, since we know the final velocity of the runner, we can find how much KE the runner gained from rest using the formula

KE= 1/2 mv^2

This is equivalent to the amount of work that the runner does. So we know W.

Now calculate power using the equation, P = W/t
 
so i got acc: 2.13 and vf: 20.64 and if mad=1/2mv^2 then i can cancel out m and I am left with 213. is 213 work?
 
Last edited:
Here's an idea. Draw a graph of velocity vs. time. The area under the curve is equal to the distance traveled. Assume that at t=0 v=0, So you find that:

v_{max} = \frac{ 2 \cdot 100 \normaltext{m} }{9.69 \normaltext{s} }

Then you should be able to use the work energy theorem to find the total work done.

Once you find the total work done, since the force is constant we must have

P = \frac{ \Delta W }{ \Delta t}
 
You need the mass to solve the problem. The problem cannot be independent of the mass. It simply takes more work to get a fat man moving fast in a given amount of time than it does to get a skinny man moving at the same speed in the same amount of time.
 
does anyone know where the formula d=1/2ma^2 comes from? or if it's even true? because then i could find the mass
 
starji said:
does anyone know where the formula d=1/2ma^2 comes from? or if it's even true? because then i could find the mass

Its not true. A good way to analyze if an equation is possibly true is to looks at its dimensions. For that equation the dimensions are

Length = Mass * Length^2 / Time^4

If the dimensions on both sides are not the same, then the equation doesn't work.

The mass should be given in the problem. The answer must depend on the mass.
 
You can use 89 kg :D Thats the average man's weight according to wolfram alpha.
 
the mass isn't given it just says an Olympic runner there has to be a away to find it without the mass
 
  • #10
these r the "challenging" questions given by my teacher so...
 
  • #11
Let me ask you this. In a given amount of time, which will it take more work to get moving at a speed of 20 m/s. A fly or a building? Can the answer be independent of the mass?
 
  • #12
thats what I am having trouble with. i know it can't be independent of the mass i just need to find a way to find the mass
 
  • #13
and where did you get d=md^2/t^4?
 
  • #14
starji said:
and where did you get d=md^2/t^4?

the dimensions of acceleration is Length / Time^2.

So the dimensions of acceleration^2 is Length^2 / Time^4

The dimension of d is Length

the dimension of m is mass
 
  • #15
starji said:
thats what I am having trouble with. i know it can't be independent of the mass i just need to find a way to find the mass

There is not enough information given to find the mass.
 
  • #16
ok i give up then i'll just ask my teacher he probably forgot
 
  • #17
starji said:
ok i give up then i'll just ask my teacher he probably forgot

Yes he probably did :biggrin:
 
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