# Homework Help: Sprinter Work and Power

1. Jun 18, 2008

### habibclan

1. The problem statement, all variables and given/known data
A 50 kg sprinter, starting from rest, runs 50 m in 7.0 s at constant acceleration.
a. What is the magnitude of the horizontal force acting on the sprinter?
b. What is the sprinter's power output at 2.0 s, 4.0 s and 6.0 s?

2. Relevant equations

xf= xo + vt+ 0.5*a*t^2
F= ma

vf= vi + at
P= (delta Work)/(Delta t)
P= f*d*cos angle

3. The attempt at a solution

Using the distance, initial velocity, and time I solve for acceleration

xf= xo + vt+ 0.5*a*t^2
50 = 0 + 0+ 0.5*a*(7^2)
a= 2.04 m/s^2

Therefore, F=ma=50 (2.04)=102 N

This, I got right. But when I calculate the power, I'm not getting the right answer when i use the first equation.
I calculate the velocity at t= 2s, and I get
vf = vi + at
vf= 2.04 ( 2)
vf= 4.08 m/s

Therefore work=delta kinetic energy= 0.5 m (vf^2 - vi^2)
K= 0.5 ( 50) ( 4.08^2)
K= 416.16 J

Therefore delta work=delta kinetic energy=416.16 J

Therefore Power= 416.16/ 2 s = 208.08 W.

Apparently the answer is just 416.16 J. When I use the second equation of P= F*v=102*4.08=416 J, I get the correct answer. Why are these two equations giving me different answers?

2. Jun 18, 2008

### Dick

Taking the difference in the kinetic energies at t=0 and t=2 and dividing by 2 sec doesn't work for the same reason that taking the difference in the positions at t=0 and t=2 and dividing by 2 sec doesn't give you the velocity at t=0. They want an 'instantaneous power' not an 'average power'.

3. Jun 18, 2008

### habibclan

I didn't know that there was a difference between the two. So whenever I have a velocity I should use the second one?

4. Jun 18, 2008

### Dick

There's definitely a difference between average and instantaneous. If they don't ask for average then they probably mean instantaneous.

5. Jun 18, 2008

### habibclan

Thanks a lot! I understand it now =).