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I am trying to determine the velocity v(t) of a person riding a motorcycle on level ground as they accelerate from zero to 15 meters/second. The person and motorcycle have a combined mass (m) kg. I assume energy is transferred from the engine into kinetic energy at a constant rate of (Pwr) watts, and that seems to me like a realistic assumption. I then get the following:

Pwr = KineticEnergy' (t)

We all know KineticEnergy(t) = (m * v(t)^2) / 2.

so we have

Pwr = m * v(t) * v' (t)

The initial condition is v(0)=0, so solution to the above differential equation is

v(t) = Sqrt( 2 * Pwr * t / m )

However, that would mean v'(0) is infinite. An infinite acceleration we would require infinite force at t=0. This is a paradox. What did I do wrong in the above derivation, and how do we get a realistic solution for v(t) in the above problem.

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1 Answer 1

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The problem is assuming constant rate of change of kinetic energy, this is not possible starting from v=0. You can see that directly from power = force times velocity, so when velocity = 0, any finite force produces zero power at first.

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  • $\begingroup$ I didn't remember (power = force times velocity). I am an electrical engineer that finished by BS degree 31 years ago. Would a good approach have us start at rest with an initial force, and assume constant (acceleration = force/mass) for a short time. Then, soon after motion begins we assume constant power? $\endgroup$
    – Ted Ersek
    Commented Mar 17, 2018 at 1:47
  • $\begingroup$ Actually power != force time velocity. They don't even have the same units. Perhaps you were in to much of a rush. You did help me see the problem with my derivation. The differential equation that I derived is (Power = MassVelocityAcceleration). Clearly if mass and power are constant in that differential equation, acceleration must approach infinity as velocity approaches zero. $\endgroup$
    – Ted Ersek
    Commented Mar 17, 2018 at 2:25
  • $\begingroup$ @Ted Ersek: Actually, force times velocity is force times (distance over time) equals (force times distance) over time equals work over time equals power... And you say exactly the same thing in your comment's fifth sentence, since Mass times acceleration equals force... $\endgroup$
    – DJohnM
    Commented Mar 17, 2018 at 3:01
  • $\begingroup$ Yes, force times velocity is certainly the power. So your question is how to realistically treat the acceleration from an engine, and honestly I don't know, but my guess is that your last suggestion is reasonable-- at first take the force to be constant, and then it should reach some maximum power, at which point the power remains constant after that. But eventually you will have air resistance, and then a maximum speed is reached where the maximum power is going into the force of the air resistance times the velocity. $\endgroup$
    – Ken G
    Commented Mar 17, 2018 at 3:18
  • $\begingroup$ @Ken G, I was wrong. Power equals force times velocity. $\endgroup$
    – Ted Ersek
    Commented Mar 18, 2018 at 17:35

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