- #1

James Brady

- 105

- 4

From Newton's 2nd Law: dv/dt = F/M.

Making this relativistic: dv/dt = [itex]\frac{F}{M}[/itex] *[itex]\sqrt{1 - v^2/c^2}[/itex]

Setting up the differential equation:

[itex]\frac{dv}{\sqrt{1 - v^2/c^2}}[/itex] = [itex]\frac{F}{M}[/itex]*dt

I multiply both sides of the equation by 1/c to make it integrateable.

[itex]\frac{dv}{C * \sqrt{1 - v^2/c^2}}[/itex] = [itex]\frac{F}{C * M}[/itex]*dt

And solving for the integral of both sides I get...

sin

^{^-1}(v/c) = [itex]\frac{F * t}{M*C}[/itex]

v(t) = sin([itex]\frac{F * t}{M*C}[/itex]) * C

I'm pretty happy with this answer , the units check out alright, and it looks good. But once you get close to the speed of light, the velocity turns around and starts going back down because sine is one of those cyclic functions. I know of course you will not all of a sudden reverse acceleration once you near the speed of light though, so I know this isn't right.

Any help would much be appreciated, I will give you an e-high-five. Or an e-hug if you're into that sort of thing. Seriously, I'm that lonely.