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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?2012-12-22
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    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.2012-12-22
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    In that case, what are you trying to do (in terms of equations)?2012-12-22
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    Solve for time taken between v=75 and 175 km/hr2012-12-22
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    Well yes, that clearly is your question, but what approach are you trying, because I don't think this is very clear from what you wrote.2012-12-22
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    I'm assuming that I should integrate this twice?2012-12-22
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    Integrate *what* twice? Please clarify your question to get more useful comment and help.2012-12-22
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    Okay, I have a differential equation in the form: C/(dx/dt) -4750 -k(dx/dt)^2 = m(d^2x/dt^2) and I need to solve for t from dx/dt =75 50 dx/dt=175. I don't know of any method to solve this.2012-12-22
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    It would be better to edit your question rather than try to explain it in comments.2012-12-22
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    And if you're editing anyway, use MathJax: http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference2012-12-22
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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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