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Here is the problem: A 500 kg car accelerates from rest to 100 m/s over a distance of 400m with the average frictional force of 1200N. If it took the car 7.3 seconds to do this, what is the power output in kW?

This was a question on Khan Academy. This is how they solved it:

​$KE=250kg*(110m/s)^2=3,025,000J$

$Work_{friction}=1200N*400m=480,000J $

$Power = \frac{3,025,000J+480,000J}{7.3 sec*1000} = 480.137 kW$

This is what I did:

$a=\frac vt$

$\Sigma F=m*v/t$

$F_{car}-F_{friction}=m*v/t$

$F_{car}=m*v/t+F_{friction}$

$Power=\frac{400(mv/t+F_{friction})}{7.3s * 1000}=478.59 kW$

What explains the difference in the answers I got? ​​

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    $\begingroup$ In the khan academy solution, it should be (100m/s)^2 instead of (110m/s)^2 $\endgroup$ – Harmohit Singh Jun 20 '17 at 16:07
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    $\begingroup$ The car's acceleration cannot have been constant. There is no constant value for the acceleration that makes the car reach 100 m/s in 7.3 s and travel 400 m. $\endgroup$ – John Rennie Jun 20 '17 at 16:13
  • $\begingroup$ I don't understand how you got your value there. I got ~441 kW; so I think you may have plugged a value in wrong. $\endgroup$ – JMac Jun 20 '17 at 16:36
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Your approach is fine. It's the given values that are defective. If you calculate the distance required for a car to reach 110 m/s in 7.3 seconds, it's actually 401.5 meters. If you insert that value into your power formula instead of the given 400 meters, you'll get the same answer they got.

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    $\begingroup$ The question gives four values: initial velocity, final velocity, distance and time. How did you decide to ignore distance? $\endgroup$ – DJohnM Jun 20 '17 at 19:51
  • $\begingroup$ My point is that the 4 given values are mutually inconsistent. I didn't ignore distance; I recalculated it assuming that the given values for velocity, time, and acceleration were precise. A short answer to the original question is that the difference between the two solutions is because of a rounding error in the problem statement. The problem writer tried to make it easier for the student by giving more information than was required, but actually introduced confusion by giving inconsistent values. $\endgroup$ – Richard H Downey Jun 21 '17 at 19:07

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