# Is it possible to model a locomotive capable of going 0-60 in 60 secs with $\sum F=F_{engine}+\mu_{rolling}mgv$?

The problem of just considering engine force and rolling friction $$\sum F=ma=F_{engine}+\mu_{rolling}mgv$$ is that when you solve for velocity like $$\dot v=\frac Fm-\mu gv$$ with initial condition $$v(0)=0$$ and get $$v(t)=\frac{F}{\mu mg}\left(1-e^{-\frac{t}{\tau}}\right)$$ and time constant, $$\tau=\frac{1}{g\mu}$$, you find out that with a steel-on-steel rolling coefficient of $$\mu=0.02$$ and a gravitational acceleration $$g=9.81m/s^2$$, in four time constant's time, or $$4\tau=0.7848$$ secs, $$v(t)$$ will settle to an equilibrium velocity.

But realistically a locomotive can only go from 0 to 60mph in 60 secs. Is it possible to model such a locomotive with my equations? If not what other factors besides rolling friction and the engine should I consider?

• Your solution is wrong. Look at my answer to your previous question – Aaron Stevens Jul 16 at 16:02
• Is $v$ here velocity? If so, you have issues with dimensions in your expression. You are summing a force ([$kgms^{-2}$] ) with something that is of units [$kgms^{-2}*ms^{-1}$]. Edit unless you forgot to add dimensions for $\mu$ – Vangi Jul 16 at 16:12
• @AaronStevens, fixed it – El Lago Jul 16 at 16:59
• @Vangi, if you are referring to my first equation, $\mu mgv$ has the units $\frac{s}{m}\times kg \times \frac{m}{s^2} \times \frac{m}{s}$. – El Lago Jul 16 at 17:01
• Is the coefficient you report what you think it is? You report it without units, which is usually what it is. But in this case it can't be. – Aaron Stevens Jul 16 at 21:05