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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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    Right, that's in latex format now. I'm not really sure on what else to add, I just want to solve for t.2012-12-22

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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.