# Car accelerated by repulsion of two point charges

• wackyvorlon

#### wackyvorlon

At the outset, I want to explain that this is a problem I came up with myself. It's not actually homework, and I suspect it is deeply conceptually flawed in some manner that I have yet to determine.

1. Two point charges of like polarity, ## q_1 = q_2 = 1C ##, start out separated by distance ## x = 1m ##. ##q_2## is attached to a car of mass ##m=1000kg##. When released, the car is accelerated by the force repelling the two charges. Find a function ## v(t) ## which gives the speed at time t.

## Homework Equations

$$F = k \frac{q_1 q_2}{x^2}$$

$$a = \frac{F}{m}$$

$$v = a t$$

Potential Energy

$$U = k \frac{q_1 q_2}{x}$$

Kinetic Energy

$$K = \frac{1}{2} m v^2$$

## The Attempt at a Solution

This has been giving me fits. I keep ending up in circular definitions. Firstly, I approach from the perspective of energy, ## U_0 = K_f ##. The end result of that was that the final speed should be ## 3.16*10^3 \frac{m}{s} ##.

## F ## becomes: $$F = \frac{k}{x^2}$$

Inserting into Newton's second law I get:

$$a = \frac{k}{m x^2}$$

Then:

$$v = \frac{k}{m x^2} t$$

You'll notice my problem. Through some means, I have to express ## x ## in terms of ## t ##, but every idea I've had relies, ultimately, on ## x ##. Truthfully, to list the approaches I've tried in detail here would require quite some typing. I feel intuitively that there ought to be some way to solve this, but frankly I'm at a loss. Any assistance you can provide in pointing me in the right direction would be greatly appreciated.

a=k/mx2
Ok.
v=(k/mx2)t
Not ok.
x is a variable. You cannot integrate x-2 by simply multiplying by t.

You can use energy conservation to find the velocity as a function of position, but getting it as a function of time is quite tricky.

Thanks! Your help is greatly appreciated.

I've been wondering if perhaps this would be a good application of the laplace transform?