Determining power output at different time intervals

[cmkluza]

Homework Statement

A 49.0 kg sprinter, starting from rest, runs 58.0 m in 9.90 s at constant acceleration.
What is the sprinter's power output at 1.90 s , 3.00 s , and 5.70 s ?

Homework Equations

$d=v_it + \frac{1}{2}at^2$
$F=ma$
$W=Fd$
$P=\frac{W}{t}$

The Attempt at a Solution

I got the acceleration from given data:
$58=(0)(9.9) + \frac{1}{2}(a)(9.9)^2 \longrightarrow a \approx 1.184 m/s^2$
Then I got the force:
$F=(49)(1.184) \approx 58 N$
Then I tried to calculate the distance traveled at different time intervals, starting with at $t=1.90$:
$d=\frac{1}{2}(1.184)(1.9)^2 \approx 2.14$
Finally, I tried to get Work and then Power from that:
$W=Fd=(58)(2.14)=123.9 J \longrightarrow P=\frac{W}{s}=\frac{123.9}{1.9} \approx 65.2 W$
However, this is not correct, the correct answer is 130 W. What am I doing wrong here?

 

[nrqed]

Homework Statement

A 49.0 kg sprinter, starting from rest, runs 58.0 m in 9.90 s at constant acceleration.
What is the sprinter's power output at 1.90 s , 3.00 s , and 5.70 s ?

Homework Equations

$d=v_it + \frac{1}{2}at^2$
$F=ma$
$W=Fd$
$P=\frac{W}{t}$

The Attempt at a Solution

I got the acceleration from given data:
$58=(0)(9.9) + \frac{1}{2}(a)(9.9)^2 \longrightarrow a \approx 1.184 m/s^2$
Then I got the force:
$F=(49)(1.184) \approx 58 N$
Then I tried to calculate the distance traveled at different time intervals, starting with at $t=1.90$:
$d=\frac{1}{2}(1.184)(1.9)^2 \approx 2.14$
Finally, I tried to get Work and then Power from that:
$W=Fd=(58)(2.14)=123.9 J \longrightarrow P=\frac{W}{s}=\frac{123.9}{1.9} \approx 65.2 W$
However, this is not correct, the correct answer is 130 W. What am I doing wrong here?

You need to use the formula Power = F v (force times velocity) and use the velocities at the given times. (Basically, what you calculated are average power over the given time intervals, while they want the instantaneous powers at the given times)

 

[cmkluza]

You need to use the formula Power = F v (force times velocity) and use the velocities at the given times

Ah okay, so using $v_f = v_i + at$ I get $v_f = (1.184)(1.9) \approx 2.25$ and then $P = Fv = 58(2.25) \approx 130 W$ is correct. If anyone sees this, could they explain why this formula works when the previous one doesn't? Still trying to wrap my head around that.

 

[nrqed]

Ah okay, so using $v_f = v_i + at$ I get $v_f = (1.184)(1.9) \approx 2.25$ and then $P = Fv = 58(2.25) \approx 130 W$ is correct. If anyone sees this, could they explain why this formula works when the previous one doesn't? Still trying to wrap my head around that.

Think about the equation $P = F \frac{d}{t}$, this is really $P = F \frac{\Delta x}{\Delta t}$, this gives a power output between two instants separated by a time $\Delta t$ and while the object moved a distance $\Delta x$. This involved a choice of two instants so one essentially calculates an average power output. If the velocity is constant, the power obtained this way is the same as the instantaneous power at every instant because the power output is constant. if the velocity is not constant, to get the instantaneous power output at a given instant one must take a limit:

$$P_{inst} = lim_{\Delta t \rightarrow 0} ~~ F ~\frac{\Delta x}{\Delta t} = F \, v_{inst} $$

 

[cmkluza]

Think about the equation $P = F \frac{d}{t}$, this is really $P = F \frac{\Delta x}{\Delta t}$, this gives a power output between two instants separated by a time $\Delta t$ and while the object moved a distance $\Delta x$. This involved a choice of two instants so one essentially calculates an average power output. If the velocity is constant, the power obtained this way is the same as the instantaneous power at every instant because the power output is constant. if the velocity is not constant, to get the instantaneous power output at a given instant one must take a limit:

$$P_{inst} = lim_{\Delta t \rightarrow 0} ~~ F ~\frac{\Delta x}{\Delta t} = F \, v_{inst} $$

Thanks a bunch, that really clarifies it!