# Calculating the power output of car gives two different answers [closed]

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? ​​

• In the khan academy solution, it should be (100m/s)^2 instead of (110m/s)^2 – Harmohit Singh Jun 20 '17 at 16:07
• 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. – John Rennie Jun 20 '17 at 16:13
• 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. – JMac Jun 20 '17 at 16:36