Question

In Problem 21 it is observed that when the velocity of the sailboat reaches 5 m/sec, the boat begins to rise out of the water and “plane.” When this happens, the proportionality constant for the water resistance drops to b₀ = 60 N-sec/m. Now find the equation of motion of the sailboat. What is the limiting velocity of the sailboat under this wind as it is planning?

Short answer

• The limiting velocity of the sailboat is 10 m/second.
• The equation of motion is $\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{10}\mathbf{t}\mathbf{+}\frac{\mathbf{25}}{\mathbf{6}}\mathbf{(}{\mathbf{e}}^{\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{t}}\mathbf{-}\mathbf{1}\mathbf{)}$.

Step-by-step solution

Find the equation of velocity

There are two forces are acting in the direction of the boat and in the opposite direction of the boat respectively.

$$\begin{gathered} \mathrm{F}=600\mathrm{N} \\ {\mathrm{F}}_2=60\mathrm{v} \end{gathered}$$

Now

role="math" localid="1664211510034" $$\begin{gathered} \mathrm{m}\frac{\mathrm{dv}}{\mathrm{dt}}=600-60\mathrm{v} \\ \mathrm{m}\frac{\mathrm{dv}}{\mathrm{dt}}=600-60\mathrm{v} \\ \frac{\mathrm{dv}}{\mathrm{dt}}=12-1.2\mathrm{v} \\ \int \frac{\mathrm{dv}}{12-1.2\mathrm{v}}=\int \mathrm{dt}(\mathrm{Integrating}\mathrm{on}\mathrm{both}\mathrm{sides}) \\ -\frac{1}{1.2}\ln \left|12-1.2\mathrm{v}\right|=\mathrm{t}+\mathrm{C} \\ \ln \left|12-1.2\mathrm{v}\right|=-1.2\mathrm{t}-1.2\mathrm{C} \\ 12-1.2\mathrm{v}=\mathrm{k}.{\mathrm{e}}^{-1.2\mathrm{t}} \end{gathered}$$

Put v = 5, t = 0 then k = 6

$$\begin{gathered} 12-1.2\mathrm{v}=6.{\mathrm{e}}^{-1.2\mathrm{t}} \\ \mathrm{v}\left(\mathrm{t}\right)=10-5.{\mathrm{e}}^{-1.2\mathrm{t}} \end{gathered}$$

Find the value of equation of motion

$$\begin{gathered} \mathrm{x}\left(\mathrm{t}\right)=\underset{0}{\overset{\mathrm{t}}{\int }}\mathrm{v}\left(\mathrm{s}\right)\mathrm{ds} \\ \mathrm{x}\left(\mathrm{t}\right)=\underset{0}{\overset{\mathrm{t}}{\int }}10-5{\mathrm{e}}^{-1.2\mathrm{s}}\mathrm{ds} \\ ={\left[10\mathrm{s}+\frac{5{\mathrm{e}}^{-1.2\mathrm{s}}}{1.2}\right]}_0^{\mathrm{t}} \\ \mathrm{x}\left(\mathrm{t}\right)=10\mathrm{t}+\frac{25}{6}({\mathrm{e}}^{-1.2\mathrm{t}}-1) \end{gathered}$$

Hence,The equation of motion is$\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{10}\mathbf{t}\mathbf{+}\frac{\mathbf{25}}{\mathbf{6}}\mathbf{(}{\mathbf{e}}^{\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{t}}\mathbf{-}\mathbf{1}\mathbf{)}$.

Find the limiting velocity

The limiting velocity of the sailboat is

$\begin{array}{rcl}\underset{t\to \infty }{\lim }v\left(t\right)& =& \underset{t\to \infty }{\lim }\left(10-5^{-1.2t}\right)\\ & =& 10\end{array}$

Hence, the Limiting velocity is 10m/sec.