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I'm stuck on this question. Originally part of a mechanics question concerning a trains's motion.

I'm finding the time taken for a train to go from $75\textrm{km/hr}$ to $175\textrm{km/hr}$

The train weighs $300T$, $\textrm{Tractive Effort}= C/v$ in $N$

$\textrm{Resistance} = 4750+kv^2$

Where $C= 2.60M$ and $k=13.3$

$T-R=ma$

$C/v-4750-kv^2=ma$

$C/(\frac{\mathrm dx}{\mathrm dt}) -4750 -k(\frac{\mathrm dx}{\mathrm dt})^2 = m\frac{\mathrm d^2x}{\mathrm dx^2}$

I need to solve for $t$ from $\frac{\mathrm dx}{\mathrm dt}=75$ to $\frac{\mathrm dx}{\mathrm dt}=175$. How do I solve for $t$?

Help appreciated!

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So... what exactly do you expect from us? Just the answer? Or is this homework, and do you want a hint? –  akkkk Dec 22 '12 at 15:14
    
Method on how to proceed, If I can't solve it with the maths that I know, then I'll have to resort to a numerical method. It's not homework, just practice. –  Elieh Dec 22 '12 at 15:17
    
In that case, what are you trying to do (in terms of equations)? –  akkkk Dec 22 '12 at 15:17
    
Solve for time taken between v=75 and 175 km/hr –  Elieh Dec 22 '12 at 15:18
2  
It would be better to edit your question rather than try to explain it in comments. –  Chris Godsil Dec 22 '12 at 15:28

1 Answer 1

Let us consider the equation $$ C/(\frac{\mathrm dx}{\mathrm dt}) -a_0 -k(\frac{\mathrm dx}{\mathrm dt})^2 = m\frac{\mathrm d^2x}{\mathrm dx^2}. $$ Just put $v=\frac{dx}{dt}$ and your equation is $$ \frac{C}{v}-a_0-kv^2=m\frac{dv}{dt}. $$ This reduces to the implicit solution $$ t_1-t_0=\int_{v_0}^{v_1}\frac{mvdv}{C-a_0v-kv^3}. $$ The integral can be approached numerically for your problem.

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