## Related questions with answers

When run at maximum power output, the motor in a self-propelled tank car can accelerate the full car (mass $M$ kg) along a horizontal track at $a$ m/s$^2$ . The tank is full at time zero, but the contents pour out of a hole in the bottom at rate $k$ 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 $t$ before the tank is empty?

Solution

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

$\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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