Q3.4-23E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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 are  before planing when the velocity is less than 5 m/sec and  when the velocity is above 5 m/sec. For sailboat B the proportionality constants are  before planing when the velocity is less than 6 m/sec and  when 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?

Step-by-Step Solution

Answer

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

Step 1: Find the equation of motion for boat A

 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.

Step 2: the equation of velocity for b 2 = 60 .

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

Step 3: the value of equation of motion boat B when B 3 = 100N .

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

Step 4: when b 4 = 50N the equation of motion for boat B

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.