1. ## 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?

2. 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.

3. 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 $30e^{10k} = 20$;

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

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

$k = 0.1 \ln \left(\frac{2}{3}\right)$

4. Thanks! how do I find out how far the boat coasts?

5. 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.