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How can constant power produce constant acceleration?

  1. Jun 6, 2015 #1
    The way I see it, a constant supply of power, say from the engine in a car to its wheels, will cause it to accelerate at a decreasing rate because the kinetic energy the power supplies is proportional to the velocity squared. That makes sense because going faster will cause the car to encounter greater resistive forces, and it's mathematically consistent (i.e. power = force*velocity or acceleration*mass*velocity, for the power to stay constant, acceleration must reduce as velocity increases) so it's a safe conclusion that constant power does not correspond to constant acceleration.

    But what if there were no resistive forces, say you have a spacecraft with a traditional chemical rocket motor in a vacuum, the rocket is at full throttle and is combusting x litres of fuel per second producing y joules of energy per second (i.e. constant power), and it has a more or less constant thrust in a vacuum, so the acceleration is constant (I know it would technically increase as the fuel tanks get lighter, but I don't think that's relevant as it is conceivable that a very efficient rocket could make this effect negligible). It seems to me that in this case, a constant acceleration is achieved from a constant power input. Anyone know what would actually happen?
     
  2. jcsd
  3. Jun 6, 2015 #2
    P = dW/dt = constant -> W = Pt + some constant that we're going to ignore for now, all work goes to KE outer space when you'll away from gravitational field, so Pt = mv^2/2, solve for v, v = sqrt(2Pt/m), whose derivative isn't constant, and so a constant power won't necessary evoque a constant acceleration,
     
  4. Jun 6, 2015 #3

    vanhees71

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    2016 Award

    Well, let's calculate it, assuming one-dimensional motion (i.e. the motion along a straight line, say the ##x## axis). First we have Newton's Equation of Motion
    $$m \ddot{x}=F.$$
    Now have to work in the assumption of constant power:
    $$m \ddot{x} \dot{x}=F \dot{x}=P=\text{const}.$$
    Now we can integrate this equation one wrt. time, because
    $$m \ddot{x} \dot{x}=\frac{m}{2} \frac{\mathrm{d}}{\mathrm{d} t} \dot{x}^2=P \; \Rightarrow \; \frac{m}{2} \dot{x}^2=P t+E_0$$
    where ##E_0=m \dot{x}_0^2/2## is the kinetic energy at ##t=0##.
    We can also go further and solve for ##x##:
    \begin{equation}
    \label{1}
    \dot{x}=\pm \sqrt{2(Pt+E_0)/m}.
    \end{equation}
    Integrating leads to
    $$x(t)=\pm \frac{2 \sqrt{2} m}{3P} \left (\frac{Pt+E_0}{m} \right )^{3/2}+x_0.$$
    The sign is determined by the direction of the motion, i.e., towards positive or negative ##x## values. For convenience let's discuss the motion with ##+##.

    The accelaration is calculated simply by taking the time derivative of (\ref{1}):
    $$\ddot{x}=\frac{P}{\sqrt{2m(P t+E_0)}}.$$
     
  5. Jun 6, 2015 #4

    russ_watters

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    Staff: Mentor

    That analysis contains an improper logical connection. You first say that power is proportional to velocity squared - which is true on its own. Then you say going faster will change the forces - which is also true, but totally separate.

    Power is proportional to velocity squared. Period. Even if you don't change the force, because the two parts are entirely different issues/parts of the equation.

    So then the real question is why is power proportional to velocity squared? Well, what is power? Work per unit time. What is work? Force times distance. So in an accelerating car, the distance per unit time is getting larger because the car is accelerating. Hence, power is getting larger.

    So, for example, if the car accelerates from a stop at 1m/s2, in the first second it travels half a meter and is moving at 1m/s at the end of the second. In the second second it is moving at 2m/s and covers an additional 1.5 m; 3x the distance in the same time, so it does 3x as much work.
     
  6. Jun 6, 2015 #5
    Good point, I got confused, I was thinking of power = force*velocity, where it can be a part of the equation, the force is the resultant force, so it's the forward thrust minus the resistive force.
     
  7. Jun 6, 2015 #6

    A.T.

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    Why velocity squared? For constant applied force power is proportional to velocity.

    "Getting larger" doesn't necessarily mean "proportional to square".
     
  8. Jun 6, 2015 #7

    Dale

    Staff: Mentor

    This is essentially correct, with the small corrections mentioned above. An automobile that delivers constant power to the wheels (over some range of speeds) will have an acceleration that decreases with increasing speed. However, the decrease is not due to increased resistive forces and would apply for idealized cases as well.

    The constant acceleration of a rocket under constant thrust is a different issue than the absence of resistive forces. In a rocket you always have to consider the exhaust. At low speeds most of the energy released in the combustion goes into accelerating the exhaust and not into accelerating the payload. As the rocket goes faster, more of the energy goes into the payload and less goes into the exhaust. So it is essentially the efficiency which is changing. A rocket has a huge amount of power, which is mostly thrown away in the exhaust at low speeds, and its efficiency to the payload increases as speed increases so that a constant acceleration is maintained.
     
    Last edited: Jun 6, 2015
  9. Jun 8, 2015 #8
    Yeah, a rocket is kind of a special and confusing case, in that on the surface it seems that the power is the same and so is the acceleration.

    There are other things to consider with rockets though, and various ways to look at them.

    Let's look at the frame of reference we're using. The faster a car goes the harder it is to accelerate...but faster relative to what? Well, that's easy: the ground that it's pushing against. But how about a rocket? It's not pushing against the ground, it's pushing against it's own fuel, which it takes with it. Essentially, it takes its frame of reference with it, so you can sort of say that its velocity is always 0, and hence the power need not increase to maintain acceleration. I don't know if that mathematically holds water, but it's one way I look at it.

    Lets freeze the frame of reference to the actual Earth though. Then we still must consider that the rocket pushes itself off its own fuel, which surely has some repercussions. In the rocket's case, the chemical energy of the fuel is not enough to tell you the whole story. You must also consider that as the rocket's speed increases, so does the KINETIC energy in it's own fuel, and this increases in kinetic energy is what allows it to accelerate constantly despite its speed increasing. The chemical energy AND the kinetic energy added together gives you a resulting total power that does in fact square with the rocket's speed (I never did the math, but I'm sure that's the case). At space rockets' speeds, this kinetic energy of the fuel actually dwarfs the fuel's chemical energy. This is called the Oberth effect, by the way.

    There's other ways to look at it, such as the efficiency changing (a rocket sitting on the launch pad with full throttle is 0% efficient). But in the end, mathematically it all comes down to the same thing: power to accelerate is a factor of velocity squared.
     
  10. Jun 8, 2015 #9

    rcgldr

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    If you consider the total mechanical energy gained by both the exhaust fuel and the rocket (ignoring losses to heat), then the total power output by the engine at some fixed throttle setting is constant from any inertial frame of reference. The rate of conversion of chemical potential energy into kinetic energy of fuel and rocket would be constant.
     
  11. Jul 27, 2016 #10
    As I see it, one must distinguish between the mechanical power which is always defined as force times velocity, and actual chemical power the engine or rocket is expending. These are not always the same. In a rocket the chemical energy is released in the rockets frame which constantly changes. It can be constant and produce a constant force and thus a constant thrust. All frames agree the same burn gives the same delta V always. As stated in the above posts, the rocket exhaust allows this state of affairs. Yet in the earth frame the mechanical power grows as thrust times velocity.

    In the case of the car I believe it is different since the force is applied in the earth frame so it takes an increasing chemical energy release to maintain the constant acceleration. This may be because the exhaust in the car case is effectively infinite, being the earth.
     
    Last edited: Jul 27, 2016
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