195. In a motorboat with an initial speed of $v_{0}=$ $=5 \mathrm{m} / \mathrm{s}$, the engine was turned off. During movement, the boat experiences water resistance, the force of which is proportional to the square of the boat's speed, with a proportionality coefficient of $m / 50$, where $m$ is the mass of the boat. After what time interval will the boat's speed decrease by half, and what distance will the boat travel during this time?
Solution. $1^{0}$. According to the condition, we can take the function as the path $s$, and the argument as time $t$. Using Newton's second law $F=m a$, we obtain the equation

$$
m s_{(t)}^{\prime \prime}=-\frac{m}{50}\left(s_{(t)}^{\prime}\right)^{2}, \text { or } s_{(t)}^{\prime \prime} \doteq-\frac{1}{50}\left(s_{(t)}^{\prime}\right)^{2}
$$

where the minus sign indicates that the resistance force is directed opposite to the motion.

$2^{0}$. This is a second-order differential equation. We reduce its order. Since $s_{(t)}^{\prime}=v_{(t)}$ and $s^{\prime \prime}(t)=v^{\prime}(t)$, we arrive at the first-order equation

$$
\frac{d v}{d t}=-\frac{1}{50} v^{2}
$$

We separate the variables and integrate:

$$
\frac{d v}{v^{2}}=-\frac{1}{50} d t ; \frac{1}{v}=\frac{t}{50}+C_{1}
$$

$3^{0}$. Using the condition $v_{0}=5$ m $/ \mathrm{c}$, we find $C_{1}=\frac{1}{5}$, from which $v=$ $=\frac{50}{t+10} \cdot$ Then we get

$$
\frac{d s}{d t}=\frac{50}{t+10} ; d s=\frac{50 d t}{t+10} ; s=50 \ln (t+10)+C_{2}
$$

From the condition $s=0$ at $t=0$ we find $C_{2}=-50 \ln 10$, from which

$$
s=50 \ln \frac{t+10}{10}
$$

We need to determine the time interval during which the speed will decrease by half. Substituting into the formula for $v$ the value $\frac{1}{2} v_{0}=2.5$, we find $2.5=\frac{50}{t+10} ; t=10$ s. During this time, the boat will travel the distance $s=50 \ln 2 \approx 34.5$ (m).