homework and exercises - Kinematics with non constant acceleration

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?

Answer

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

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

or

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

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.