Sailboats A and B each have a mass of 60 kg and cross the starting line at the same time on the first leg of a race. Each has an initial velocity of 2 m/sec. The wind applies a constant force of 650 N to each boat, and the force due to water resistance is proportional to the velocity of the boat. For sailboat A the proportionality constants arebefore planing when the velocity is less than 5 m/sec andwhen the velocity is above 5 m/sec. For sailboat B the proportionality constants arebefore planing when the velocity is less than 6 m/sec andwhen the velocity is above. If the first leg of the race is 500 m long, which sailboat will be leading at the end of the first leg?

There are two forces are acting on the sailboat are the wind force and the water resistance force respectively.

$$\begin{gathered} F=650N \\ {F}_2=80v \end{gathered}$$

Now put the given values then;

$$\begin{gathered} m\frac{dv}{dt}=650-80v \\ 60\frac{dv}{dt}=650-80v\pm \\ \frac{dv}{dt}=\frac{65}{6}-\frac{4}{3}{v}_A \\ \frac{dv}{dt}+\frac{4}{3}{v}_A=\frac{65}{6} \\ {v}_A=\frac{65}{6}+C{e}^{-\frac{4}{3}t}\left(\mathbf{Integrating}\mathbf{}\mathbf{on}\mathbf{}\mathbf{both}\mathbf{}\mathbf{sides}\right) \end{gathered}$$

Put the value of $v_A\left(0\right)=2, then C=-\frac{49}{8}$.

$$\begin{gathered} v_A\left(t\right)=\frac{65}{6}-\frac{49}{8}{e}^{-\frac{4}{3}t} \\ x \left(t\right)=\underset{0}{\overset{t}{\int }}v\left(s\right)ds \\ x \left(t\right)=\underset{0}{\overset{t}{\int }}\left(\frac{65}{8}-\frac{49}{8}{e}^{-\frac{4}{3}s}\right)ds \\ x \left(t\right)=\frac{65}{8}t-\frac{147}{32}\left(e^{-\frac{4}{3}t}-1\right) \end{gathered}$$

Further, solve the above expression,

$$\begin{gathered} x\left(t\right)=\underset{0}{\overset{t}{\int }}\left(\frac{65}{8}-\frac{49}{8}{e}^{-\frac{4}{3}s}\right)ds \\ x\left(t\right)=\frac{65}{8}t-\frac{147}{32}\left(e^{-\frac{4}{3}t}-1\right) \end{gathered}$$

When the values are written as;

$$\begin{gathered} v_A\left(t\right)=5, \\ then, t_A=0.5 \\ {v}_A\left(t\right)=0.5, \\ then, x_A=1.85 \end{gathered}$$

Since, the first leg of race is 500 m long the boat A will be $500-1.86=498.15 m$away from the finish.

Apply the same procedure as step 1 b₂ = 60 then I get the required value.

Therefore, the value is $v_{2A}\left(t\right)=\frac{65}{6}-\frac{35}{6}{e}^{-t}$.

And the equation of motion is written as:

$$\begin{gathered} x_{2A}\left(t\right)=\frac{65}{6}t+\frac{35}{6}\left(e^{-t}-1\right) \\ {x}_{2A}\left(t\right)=\underset{0}{\overset{t}{\int }}\left(\frac{65}{6}-\frac{35}{6}{e}^{-s}\right)ds \end{gathered}$$

When the value of $x_{2A}=498.15,then, t=46.5$.

Now the total time for the boat A is = 47 sec

Apply the same procedure for step 1 the equation of velocity and equation of motion are respectively.

$$\begin{gathered} v_{1B}=\frac{65}{8}-\frac{49}{8}{e}^{-\frac{5t}{3}} \\ {x}_{1B}\left(t\right)=\frac{65}{8}t+\frac{147}{6}\left(e^{\frac{-5t}{3}}-1\right) \end{gathered}$$

The time is 0.635 sec

Apply the same procedure for step 2 the equation of velocity and equation of motion are respectively.

$$\begin{gathered} v_{2B}=\frac{65}{5}-\frac{35}{8}{e}^{-\frac{5t}{3}} \\ {x}_{2B}\left(t\right)=\frac{65}{5}t+\frac{42}{5}\left(e^{\frac{-5t}{3}}-1\right) \end{gathered}$$

The time is 39.895 sec

Therefore, the boat B will be leading at the end of the first leg.