# Why is the steady speed neglected when calculating work or power? [closed]

A driver notices that her 1080-kg car, when in neutral, slows down from 95 km/hr to 65 km/hr in about 7.0s on a flat horizontal road. Approximately what power (watts and hp) is needed to keep the car traveling at a constant 80 km/h?

Here's what I did, I calculated the energy I would need to oppose the 95-65 change and then I added the energy needed to move at 80 km/hr then I divided by 7 . I think I have a misconception + a mistake somewhere because the book's answer is different. Can someone tell me where I went wrong ?

P.s, I know I can calculate the acceleration and then use F=ma AND F=PV but I don't understand why the work isn't energy needed to oppose 95-65 + KE for V=80. If you know, please answer I have been trying to understand this for 6 hours.

The other answers are correct, I just want to try a different wording to see if it clears it up.

"Here's what I did, I calculated the energy I would need to oppose the 95-65 change and then I added the energy needed to move at 80 km/hr then I divided by 7 . "

Emphasis mine. No energy is needed for the car to keep moving at $$80 \ kph$$ if there were no resisting force. If the car is already at $$80 \ kph$$, the energy needed to maintain that speed only depends on the force resisting the movement.

The kinetic energy required is $$\Delta KE = \frac 12 m v_{1}^2 - \frac 12 m v_0^2$$ and since $$v_0 = v_1 = 80 \ kph$$, $$\Delta KE = 0$$. You only need to look at the energy required to overcome the resistance; because besides that, the object will want to maintain the $$80 \ kph$$.

• Shouldn't the fact that the the speed is 80 km/h and not some other speed matter for the power though? – Aaron Stevens Dec 8 '19 at 21:11
• @AaronStevens Yes, it will have an effect on power to oppose resisting force, but OP was saying "+ the KE for V=80" which made me think that was what confused him. I think he tried to add on the KE of an object moving 80 kph; but that doesn't make sense when you consider that it's already at that speed. – JMac Dec 8 '19 at 21:38
• Right. The weird thing is that the answer by TheLastAirbender has the answer as $\Delta K/t$ for the 95 km/h to the 65 km/h over the 7 seconds, which does give the right answer without even considering the 80 km/h... I think it might just be a coincidence. I don't think in general that the average power during the slow down is equal to the power to maintain any constant speed. – Aaron Stevens Dec 8 '19 at 21:42
• I don't think so. Their answer says the power is equal to $\Delta K/t$ – Aaron Stevens Dec 8 '19 at 21:44
• @AaronStevens Nevermind I think you're right about that. It seems pretty strange here that it gets the right answer, but I think I see why. Because the average of 95 and 65 is 80, the average change in KE over the 7 second window would be exactly the same as the power required to maintain the speed at 80 kph with that dissipation. I'm not sure if averaging the dissipative force like that actually makes sense though... but I had to make that assumption when solving in the P = Fv method anyways. – JMac Dec 8 '19 at 21:56