I am interested in trying to understand the meaning of the integral of power with respect to distance.

In the case of force, we have the two formulae: $\int F \, \operatorname{d}\!t = \Delta p$ and $\int F \, \operatorname{d}\!x = \Delta E_k$.

I understand that power is closely linked to time and work by the relationship $P = \frac{\operatorname{d}\!W}{\operatorname{d}\!t}$.

However, imagine a body moving continuously from the left to the right. In such a case, we can express displacement as a one-to-one function of time, and time as a one-to-one function of displacement. We could place regular markers along the road side and take power readings as the body passes each of these markers. Assuming the distance between markers is $\Delta s$ and then letting $\Delta s \to 0$ gives power as a function of displacement. We may then ask: What does the integral of this function mean?

$$\int_{s_1}^{s_2} P \, \operatorname{d}\!x = \operatorname{F}(s_1,s_2)$$


I've tried to play with the calculus a bit. I'm not sure if this helps:

$$P = \frac{\operatorname{d}\!W}{\operatorname{d}\!t} = \frac{\operatorname{d}\!x}{\operatorname{d}\!t} \frac{\operatorname{d}\!W}{\operatorname{d}\!x} = v\frac{\operatorname{d}\!W}{\operatorname{d}\!x}\implies \int P \, \operatorname{d}\!x = \int v\frac{\operatorname{d}\!W}{\operatorname{d}\!x} \, \operatorname{d}\!x = \int v \, \operatorname{d}\!W$$

Has anyone seen a velocity-work graph? If so, what does the area underneath it represent?

  • $\begingroup$ Related question by OP: physics.stackexchange.com/q/77897/2451 $\endgroup$ – Qmechanic Sep 22 '13 at 15:03
  • $\begingroup$ @Qmechanic That question has been flagged for moderator deletion for some time. It contained an error that neither you nor I spotted. I was unable to delete because of your own, much appreciated, reply. Could you please delete the question? $\endgroup$ – Fly by Night Sep 22 '13 at 16:50
  • $\begingroup$ It seems to be a good question, so there's really no reason to delete it. $\endgroup$ – David Z Nov 21 '13 at 22:30
  • $\begingroup$ The result is a vector quantity since ${\rm d}x$ is vector really (you mentioned direction anyway). $\endgroup$ – ja72 Feb 20 '14 at 5:55

Maybe this can help. Consider a body accelerating under constant power from the initial conditions $t=0$, $x=x_1$ and $v=v_1$.

$$ a(v) = \frac{P}{m v} $$

$$ t(v) = \int \frac{1}{a(v)}\,{\rm d} v = \int \frac{m v}{P} \,{\rm d}v = \frac{m}{P} \left( \frac{v^2}{2} - \frac{v_1^2}{2} \right) $$

$$ v(t) = \sqrt{ \frac{2 P t}{m} + v_1^2} $$

$$ x(v) = \int \frac{v}{a(v)}\,{\rm d} v = \int \frac{m v^2}{P} \,{\rm d}v =x_1 + \frac{m}{P} \left( \frac{v^3}{3} - \frac{v_1^3}{3} \right)$$

$$ v(x) = \sqrt[3]{\frac{3 P x}{m} + v_1^3} $$

$$ x(t) = x(v(t)) =x_1 + \frac{m}{P} \left(\frac{1}{3} \left( \frac{2 P t}{m} + v_1^2 \right)^{\frac{3}{2}} - \frac{v_1^3}{3} \right) $$

So it seems the quantity you are looking for is

$$ \int P \,{\rm d}x= P \int {\rm d}x=P\, x(v) $$ for constant power.

In general, power is a function of speed, so you can say

$$ \int P(v) \,{\rm d}x = \int P(v) \frac{v}{a} \,{\rm d}v = \int m v^2 \,{\rm d}v = \frac{1}{3} m \left( v^3 - v_1^2 \right) $$

which appears to be scaling factor for $x(v)$. In fact, you can state that the average power within a distance is

$$ P_{ave} = \frac{ \int P \,{\rm d} x}{\int \,{\rm d} x} = \frac{ \frac{1}{3} m \left( v^3 - v_1^3 \right)}{x-x_1} $$

So you can take a measurement of the distance traveled ($x_1 \rightarrow x$) and the speed change ($v_1 \rightarrow v$) and estimate the average power needed to reach those numbers. Same way that people use the trap speed of a 1/4 mile drag race to estimate power.

| cite | improve this answer | |
  • $\begingroup$ I fixed a typo in the integrals for $t(v)$ and $x(v)$. $\endgroup$ – ja72 Feb 20 '14 at 14:04

What you did so far is correct.

Use the fact that $ W = 1/2 \cdot m \cdot v^{2}$.

Here are the final steps from your work to the solution, use integration by parts:

$$ \int v \cdot dW = 1/2 \cdot m \cdot v^{3}|_{v_1}^{v_2} - \int_{v_1}^{v_2} 1/2 \cdot m \cdot v^{2} =1/3 \cdot m \cdot [v_2^3 - v_1^3] $$

| cite | improve this answer | |
  • $\begingroup$ That's very interesting. I arrived at the formula $\frac{1}{3}mv^3$ myself by integrating $P = Fv$ directly. The problem was that $P=Fv$ only holds for constant $v$. I think that $W = \frac{1}{2}m(v^2-u^2)$, and not just $\frac{1}{2}mv^2$. I would also like to understand what the answer means. $\endgroup$ – Fly by Night Sep 23 '13 at 18:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.