5. The force with which the airflow acts on the sail can be calculated using the formula

$F=\frac{A S \rho\left(v_{0}-v\right)^{2}}{2}$, where $A-$ is the aerodynamic force coefficient, $S-$ is the area of the sail $S$ $=4 \mathrm{m}^{2} ; \rho$ - density of air, $v_{0}$ - wind speed $v_{0}=4.8 \mu / c, v$ - speed of the sailboat. At some moment in time, the instantaneous power of the wind reaches its maximum value. What is the speed of the sailboat at this moment?

Given:

$F=\frac{A S \rho\left(v_{0}-v\right)^{2}}{2}$

$N\left(t^{\prime}\right)=N_{\max }$

$v_{0}=4.8 \frac{M}{c}$

$v\left(t^{\prime}\right)-?$
Solution:

$$
\begin{aligned}
& \left.\begin{array}{l}
F=f(v) \\
N=F \cdot v
\end{array}\right\} \Rightarrow N=f(v) \\
& N=\frac{A S \rho}{2}\left(v_{0}^{2}-2 v_{0} v+v^{2}\right) v=\frac{A S \rho}{2}\left(v_{0}^{2} v-2 v_{0} v^{2}+v^{3}\right) \\
& N=N_{\max } \Rightarrow N^{\prime}(v)=0 \\
& N^{\prime}(v)=\frac{A S \rho}{2}\left(v_{0}^{2}-4 v_{0} v+3 v^{2}\right)=0 \\
& v_{0}^{2}-4 v_{0} v+3 v^{2}=0 \\
& 3 v^{2}-4 v_{0} v+v_{0}^{2}=0
\end{aligned}
$$

$$
\begin{aligned}
& v_{1,2}=\frac{4 v_{0} \pm \sqrt{16 v_{0}^{2}-12 v_{0}^{2}}}{6}=\frac{4 v_{0} \pm \sqrt{4 v_{0}^{2}}}{6}=\frac{4 v_{0} \pm 2 v_{0}}{6} \\
& v_{1}=v_{0} \text{ (not suitable)} \quad v_{2}=\frac{v_{0}}{3}=1.6 \text{ m/s}
\end{aligned}
$$

## OPTION 3