A particle experiences an acceleration described by $$ a=kx^{-2} $$ where x is the displacement from the origin and k is an arbitrary constant.

To what value does the velocity v of the particle converge to as x approaches infinity if the particle starts at some point x0?

If I approach this problem with energy, then $$ W = \int F \mathrm{d}x $$ $$ = \int_{x_0}^\infty mkx^{-2} \mathrm{d}x $$ $$ = mk(-\infty^{-1}+{x_0}^{-1}) $$ $$ W = K = mk{x_0}^{-1} $$ $$ \frac{1}{2}mv^2 = mk{x_0}^{-1} $$ $$ v = \sqrt{2k{x_0}^{-1}} $$

How would I solve this problem with pure kinematics? (there appears to be some sort of cyclical dependency where acceleration affects velocity, velocity affects displacement, and displacement affects acceleration)

Likewise, two particles experience accelerations described by $$ a_1=k_1x^{-2} $$ $$and $$ $$ a_2=-k_2x^{-2} $$ where x is the distance between the two particles

What two velocities do the particles reach as x approaches infinity if the two particles are initially separated by some x0?


1 Answer 1


The solve the first part with just kinematics, use the chain rule:

$$ a(x) = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = \frac{dv}{dx}v,$$ and then $a(x)dx = vdv$.

Integrating both sides ($x_0$ to infinity on the left and $v_0$ to $v_f$ on the right), we get

$$\frac{k}{x_0} = \frac{v_f^2 - v_0^2}{2},$$


$$v_f = \sqrt{\frac{2k}{x_0} + v_0^2}.$$

Solving the two particle scenario is no more complicated than the single particle version as long as you pay attention to signs for particle 2.

  • 1
    $\begingroup$ Note that this site uses MathJax for equation rendering. Just put the math between \$ ... \$ inline or \$\$ ... \$\$ for display equations and use standard latex formatting. $\endgroup$
    – Michael
    Apr 9, 2013 at 2:27
  • $\begingroup$ @MichaelBrown I was actually slowly in the process of trying to convert the text, but I've never tried it before today. Thanks for the help though. =) $\endgroup$
    – David H
    Apr 9, 2013 at 2:33
  • $\begingroup$ Seeing as my calculus is fairly weak, could you show how the 2 particle one would work as well? $\endgroup$
    – Andrew Liu
    Apr 9, 2013 at 3:24
  • $\begingroup$ @Andrew Don't worry, that one's easy. Use it as practice for yourself to make sure you really understand the answer. ;-) $\endgroup$
    – David Z
    Apr 9, 2013 at 4:20
  • $\begingroup$ Ok, I've concluded that the total velocity v=sqrt(2(k1+k2)/x0), is this correct? If so, could you point me in the right direction as to how to find the individual velocities, v1 and v2? $\endgroup$
    – Andrew Liu
    Apr 9, 2013 at 17:03

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