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\textbf{Vector Functions and Curves}
\textit{\textbf{Applications}}
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\textbf{Question}
If run at full power, a self-propelled tank car can accelerate itself,
when full (mass = $M$kg) along a straight track at $a$
m/se$\textrm{c}^2$.
At time $t=0$ the tank is full. The contents leave the tank at a rate
$k$ kg/sec after that.
At time $t=0$ the car is at rest and then full power is activated to
accelerate the car forwards. How fast will it be moving at any time
$t$ before the tank is empty?
\textbf{Answer}
At time $t$ seconds, the speed of the car is $v(t)$ and its mass is
$m(t)=M-kt$kg.
When the car is at full power, since the motor accelerates the car at
$a$ m/$\textrm{s}^2$ the force applied is $F=ma$.
The force is the rate of change of momentum of the car (by Newton's
Law).
$\Rightarrow$
\begin{eqnarray*}
\frac{d}{dt} \left [ (M-kt)v \right ] & = & Ma\\
(M-kt) \frac{dv}{dt} - kv & = & Ma\\
\frac{dv}{Ma+kv} = \frac{dt}{M-kt}\\
\frac{1}{k} \ln (Ma+kv) = -\frac{1}{k} \ln (M-kt) + \frac{1}{k} \ln
C\\
Ma+kv & = & \frac{C}{M-kt}
\end{eqnarray*}
When $t=0$, $v=0$, $\Rightarrow Ma=\frac{C}{M}$.
$\Rightarrow$
\begin{eqnarray*}
C & = & M^2a\\
kv & = & \frac{M^2a}{M-kt}-Ma = \frac{Makt}{M-kt}\\
\Rightarrow v(t) & = & \frac{Mat}{M-kt} \textrm{m/s}
\end{eqnarray*}
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