I asked my son to figure out power consumption of a car travelling on a horizontal at constant acceleration without friction. We know the correct answer is Force * average_velocity, but would like help understanding why the following gives the wrong answer: initial velocity and position are zero -- P - power W - work F - force d - distance a - acceleration t- time v- velocity P = W/t = F*d/t = F * a * t^2/2t = F*v/2 --- All we did is subsitute distance as at^2/2, and v/t for 'a' ...
It's a trick question: if there's no friction between the wheel and the road, there can be no acceleration. The car did, does, and will forever have the constant velocity of zero. Newton's First Law, IIRC
F=m*a . It fixes almost everything. **EDIT** I think you need to back up to the beginning. What is given in the problem statement?
Why do you say it is the wrong answer? v/2 is the average velocity so the result is Force times the average velocity as you originally stated. kevin PS I assume that your term "without friction" means without losses caused by air resistance, rolling resistance etc, not that there is no friction between the tire and the road.
Hi Kevin, I cannot convince myself that v/2 is average velocity. Seems like it should be v(final)/t ? Help please
Hi Sage, I think your problem is that a is Delta-v/Delta-t, not v/t. You are confusing the two. So, the equivalence you use is only valid if v0 is (starting velocity) zero, and startint time t0, is zero. In which case v/2 is the average velocity, if accelleration is constant, and v is the final velocity (speed actually, since velocity is a vector quantity). In otherwords your final equation should be: P = F*(v1-v0)/2 , assuming t0 is zero. since a = (v1-v0)/(t1-t0)
Thanks everybody, I finally get it. v/2 is the average speed, not v/t. I think it was solving SAT questions ad nauseum with my kids that turned me into a block of cheese. You know -- average is Sum of values/number of elements. It can still be calculated this way, but the numerator is vt/2 and the denominator is t.
