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So recently I came about this report on magnetic braking which I mostly got. However, there was a proof that they skipped over and went straight to the solution that I would like to understand.

Would it be possible to show your working to get from $mdv/dt = mg\sin\theta − bv − \mu mg\cos\theta$ to $v = vT [1 − exp(−t/\tau)]$ where $vT$ is the terminal velocity equal to $(mg\sin\theta − \mu mg\cos\theta)/b$.

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We have $m\frac{dv}{dt}=mg\sin{\theta}-bv-\mu mg\cos\theta $. So $\frac{dv}{dt}+\frac{bv}{m}=g\sin{\theta}-\mu g\cos\theta$, and we get $ve^{\int{\frac{b}{m}}dt}=\int (g\sin{\theta}-\mu g\cos\theta)(e^{\int{\frac{b}{m}}dt})dt$, so $ve^{\frac{bt}{m}}=(g\sin{\theta}-\mu g\cos\theta)e^{\frac{bt}{m}}(\frac{m}{b}) +c$. They denote $(g\sin{\theta}-\mu g\cos\theta)(\frac{m}{b})=v_T$ and, and from initial conditions when $t=0,v=0 $, so $c=-v_T$.Dividing through by $e^{\frac{bt}{m}}$, we obtain $v=v_T-v_Te^{\frac{-bt}{m}}$, which simplifies to the required integral.

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