# velocity problem, calculus?

• Oct 6th 2008, 02:24 PM
wvlilgurl
velocity problem, calculus?
A motorboat is moving at 30m/s when its motor suddenly quits and it is 800 m from the shore. Ten seconds later the boat has slowed to 20 m/s. Assume that the resistance it encounters while coasting is proportional to its velocity. How far will the boat coast in all? Can the boat reach the shore?
• Oct 6th 2008, 02:40 PM
icemanfan
If the resistance is proportional to the boat's velocity, then the deceleration of the boat is also proportional to its velocity. So you have the equation $v(t) = kv'(t)$. You have to use this equation to determine an equation for the boat's distance traveled starting at the time the boat begins to slow down.
• Oct 6th 2008, 03:04 PM
icemanfan
The equation $v(t) = kv'(t)$ is a common differential equation. The solution is $v(t) = Ce^{kt}$. Since v = 30 at t = 0, this yields $30 = Ce^{k(0)} = Ce^0 = C$, so the equation is $v(t) = 30e^{kt}$. Now, since v(10) = 20, we have $30e^{10k} = 20$;

$e^{10k} = \frac{20}{30} = \frac{2}{3}$

$10k = \ln \left(\frac{2}{3}\right)$

$k = 0.1 \ln \left(\frac{2}{3}\right)$
• Oct 6th 2008, 04:18 PM
wvlilgurl
Thanks! how do I find out how far the boat coasts?
• Oct 6th 2008, 04:37 PM
icemanfan
The equation for position will be the antiderivative of the velocity function, which is $s(t) = \frac{30}{k}e^{kt} + C$. Since s(0) = 0, $s(0) = \frac{30}{k}e^0 + C = \frac{30}{k} + C = 0$. Hence $C = -\frac{30}{k}$ and
$s(t) = \frac{30}{k}e^{kt} - \frac{30}{k}$. The limit of this function as t approaches infinity is $\lim_{t \to \infty} \left(\frac{30}{k}e^{kt} - \frac{30}{k}\right) = 0 - \frac{30}{k}$, and that is the distance the boat coasts.