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{\bf Question}

$(*)$ A car with a mass of 1000kg accelerates down a road with the
engine producing a force of 2000$kg$ $m/sec^2$. Determine how
quickly the car reach $10m/s$ from stationary and how far it
travels in this time.
\par Now consider extending the model to account for the fact that the
car is subjected to wind resistance which produces a resistance
force proportional to the speed of the car (the constant of
proportionality is $40$ (with all measurements in $m$, $kg$ and
$sec$). How much longer does it take to get to $10m/sec$ when this
air resistance force is accounted for. What is the maximum speed
of the car in this case.


\vspace{0.25in}

{\bf Answer}

$\ds 1000\frac{d^2x}{dt^2}=2000 \Rightarrow \frac{d^2x}{dt^2}=2
\Rightarrow x=t^2+At+B$

$\ds \Rightarrow x(0)=\frac{dx}{dt}(0)=0 \Rightarrow x=t^2,
\,\,\,\, \frac{dx}{dt}=2t$

$\ds \frac{dx}{dt}=10$ when $\ds 2t=10 \Rightarrow t=5$ sec, at
$\ds t=5$ sec, \,\, $\ds x=(5)^2=25$metres.

Now with air resistance $\ds
1000\frac{d^2x}{dt^2}=2000-40\frac{dx}{dt}$

The linear equation for $\ds v$ is $\ds 1000\frac{dv}{dt}+4v=200
\Rightarrow \frac{dv}{dt}+\frac{1}{25}v=2$

$\ds
I(x)=\exp\left(\int\frac{1}{25}dt\right)=\exp\left(\frac{t}{25}\right)$

$\ds\Rightarrow
\frac{d}{dt}\left(v\exp\left(\frac{t}{25}\right)\right)=2\exp\left(\frac{t}{25}\right)$

$\ds \Rightarrow ve^\frac{t}{25}=50e^\frac{t}{25}+A$

$\ds v(0)=0 \Rightarrow A=-50$

$\ds v(t)=50\left(1-e^{-\frac{t}{25}}\right)$

maximum speed is when $\ds t\rightarrow\infty$ and $\ds
v\rightarrow 50$m/s

$\ds v=10$ when $\ds 10=50\left(1-e^{-\frac{t}{25}}\right)
\Rightarrow \ln\left(1-\frac{1}{5}\right)=-\frac{1}{25}t$

$\ds t=25\ln\frac{5}{4}\approx5.578 \Rightarrow$ air resistance
slows the car so it takes 0.578 seconds longer to get to 10m/s.


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