# How far does the boat coast?

A motorboat is moving at 20 m/s when the engine quits. Assume that the resistance is proportional to the velocity. Also assume that after 20 seconds the velocity is 10 m/s. How far does the boat coast?

I need help sorting the information out. What exactly does it mean for the boats engine to have quit while it's moving 20 m/s? What does that tell us? So far I have:

$\frac{dv}{dt}=-kv$

$v(20) = 10$

Where to go from here? How do I find $k$? What am I trying to "solve" to figure out how far the boat coasts?

Thank you!

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You're right so far. You know that

$$\frac{dv}{dt} = -kv \tag{1}$$

for some $k$. Its important to note that this equation is only valid when the boat's engine is not running (otherwise there would be some other term on the right-hand side accounting for the effect of the engine).

You also have the conditions $v(0)=20$ and $v(20)=10$. You can solve (1) to find

$$v(t) = Ae^{-kt}$$

and then use your two conditions on the speed as follows

$$v(0) = A = 20$$

$$v(20) = Ae^{-20k} = 10$$

to solve for $A$ and $k$, finding $A=20$ and $k=\ln(2)/20$. The distance coasted is then given by solving

$$\frac{dx}{dt} = v \quad\Rightarrow\quad x(t) = \int_0^t v(s)ds$$

to get

$$x(\infty) = \int_0^\infty 20e^{-(s\ln 2)/20} ds$$

Why is $v(t) = Ae^{-kt}$, I get $v(t) = -k\frac{v^2}{2}$? – CodyBugstein Oct 7 '12 at 20:35
@Imray You have to integrate with respect to $t$, not with respect to $v$. Try rewriting as $dv/v = -k dt$ and integrating both sides. – Chris Taylor Oct 7 '12 at 21:26
Ah OK, so divide both sides by $v$, then multiply both sides by $dt$ and end up with $v(t) = e^{-kt}$. Where does the $A$ come from? – CodyBugstein Oct 11 '12 at 17:06
From $dv/d=-kdt$ you integrate to get $\ln(v)=-kt+c$. Now exponentiating, you get $v=e^c e^{-kt}$. We write $A=e^c$ just to make things neater. – Chris Taylor Oct 11 '12 at 17:16