Lowemack Posted January 28, 2007 Share Posted January 28, 2007 Does anybody know what the formula is for the graph you see in text books, of speed against time for a constant force applied to an object. The one where speed increases with time, and then levels of as it approaches C. Link to comment Share on other sites More sharing options...
timo Posted January 28, 2007 Share Posted January 28, 2007 I know how to figure it out: 1) Find out what the momentum after time t is, when a constant force is applied 2) Find out how the velocity of an object depends on its momentum (and its mass/energy). 3) Plug it together. Link to comment Share on other sites More sharing options...
Lowemack Posted January 28, 2007 Author Share Posted January 28, 2007 For non reletavistic speeds I think the formula is f=ma, which when re-arranged would give t=mv/f. and for a force of 1N applied to a mass of 1kg t=v. So after 10 seconds the mass would be doing 10m/s,after 20s 20m/s. But at speeds close to see we know that this is not the case, so does anybody know the formula that matches the graph I mentioned? Link to comment Share on other sites More sharing options...
Lowemack Posted January 29, 2007 Author Share Posted January 29, 2007 sorry, I meant to post this in 'Relativity'. Can anyone move it please? Link to comment Share on other sites More sharing options...
timo Posted January 31, 2007 Share Posted January 31, 2007 You should go via momentum. Sadly, no one else bothered replying here and I will probably not have time to guide you to the equation within the next weeks. So I will just give you the formula I suppose you are looking for: [math] F = \frac{dp}{dt} \Rightarrow p = Ft, v = \frac{pc^2}{E} = \frac{pc^2}{\sqrt{m^2c^4 + p^2c^2}} \Rightarrow v = \frac{Ftc^2}{\sqrt{m^2c^4 + F^2 t^2 c^2}} [/math] The difference is in part 2) of the points I previously mentioned. In classical mechanics, the relation of velocity and momentum is p = mv. In relativistic mechanics, the relation is different (the one I used above). Link to comment Share on other sites More sharing options...
Lowemack Posted January 31, 2007 Author Share Posted January 31, 2007 You should go via momentum. Sadly, no one else bothered replying here and I will probably not have time to guide you to the equation within the next weeks. So I will just give you the formula I suppose you are looking for: [math] F = \frac{dp}{dt} \Rightarrow p = Ft, v = \frac{pc^2}{E} = \frac{pc^2}{\sqrt{m^2c^4 + p^2c^2}} \Rightarrow v = \frac{Ftc^2}{\sqrt{m^2c^4 + F^2 t^2 c^2}} [/math] The difference is in part 2) of the points I previously mentioned. In classical mechanics, the relation of velocity and momentum is p = mv. In relativistic mechanics, the relation is different (the one I used above). Thanks a lot, that works perfect. Link to comment Share on other sites More sharing options...
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