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Question

When run at maximum power output, the motor in a self-propelled tank car can accelerate the full car (mass MM kg) along a horizontal track at aa m/s2^2 . The tank is full at time zero, but the contents pour out of a hole in the bottom at rate kk kg/s thereafter. If the car is at rest at time zero and full forward power is turned on at that time , how fast will it be moving at any time tt before the tank is empty?

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Answered 4 months ago
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Let v(t)v(t) be the speed of the tank car at time tt seconds. The mass of the car at time tt is m(t)=Mktkg\color{#4257b2}m(t)=M-ktkg . At full power, the force applied to the car is F=MaF=Ma (since the motor can accelerate the full car at aa m/s2s^2). By Newton's Law, this force is the rate of change of the momentum of the car. Thus

ddt[mv]=Maddt[(Mkt)v]=Ma(Mkt)dvdtkv=Ma(Mkt)dvdt=Ma+kvdvMa+kv=dtMktdvMa+kv=dtMkt1kln(Ma+kv)=1kln(Mkt)+C11kln(Ma+kv)=1kln(Mkt)+1klnCln(Ma+kv)=ln(Mkt)+lnCln(Ma+kv)=lnCMktMa+kv=CMkt\begin{align*} \dfrac{d}{dt}[mv]&=Ma\\ \dfrac{d}{dt}\Big[(M-kt)v\Big]&=Ma\\ (M-kt)\dfrac{dv}{dt}-kv&=Ma\tag{Product Rule}\\ (M-kt)\dfrac{dv}{dt}&=Ma+kv\\ \dfrac{dv}{Ma+kv}&=\dfrac{dt}{M-kt}\tag{Separate variables}\\ \int{\dfrac{dv}{Ma+kv}}&=\int{\dfrac{dt}{M-kt}}\tag{Integrate}\\ \dfrac{1}{k}\ln(Ma+kv)&=-\dfrac{1}{k}\ln(M-kt)+C_1\tag{Find antiderivative of both sides}\\ \dfrac{1}{k}\ln(Ma+kv)&=-\dfrac{1}{k}\ln(M-kt)+\dfrac{1}{k}\ln C\tag{Let $C_1=\dfrac{1}{k}\ln C$}\\ \ln(Ma+kv)&=-\ln(M-kt)+\ln C\\ \ln(Ma+kv)&=\ln\dfrac{C}{M-kt}\\ {\color{Fuchsia}Ma+kv}&{\color{Fuchsia}=\dfrac{C}{M-kt}} \end{align*}

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