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Q3.4-22E

Expert-verifiedFound in: Page 117

Book edition
9th

Author(s)
R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages
616 pages

ISBN
9780321977069

**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 b_{0} = 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?**

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

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

$\mathrm{F}=600\mathrm{N}\phantom{\rule{0ex}{0ex}}{\mathrm{F}}_{2}=60\mathrm{v}\phantom{\rule{0ex}{0ex}}$

Now

role="math" localid="1664211510034" $\mathrm{m}\frac{\mathrm{dv}}{\mathrm{dt}}=600-60\mathrm{v}\phantom{\rule{0ex}{0ex}}\mathrm{m}\frac{\mathrm{dv}}{\mathrm{dt}}=600-60\mathrm{v}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{dv}}{\mathrm{dt}}=12-1.2\mathrm{v}\phantom{\rule{0ex}{0ex}}\int \frac{\mathrm{dv}}{12-1.2\mathrm{v}}=\int \mathrm{dt}(\mathrm{Integrating}\mathrm{on}\mathrm{both}\mathrm{sides})\phantom{\rule{0ex}{0ex}}-\frac{1}{1.2}\mathrm{ln}\left|12-1.2\mathrm{v}\right|=\mathrm{t}+\mathrm{C}\phantom{\rule{0ex}{0ex}}\mathrm{ln}\left|12-1.2\mathrm{v}\right|=-1.2\mathrm{t}-1.2\mathrm{C}\phantom{\rule{0ex}{0ex}}12-1.2\mathrm{v}=\mathrm{k}.{\mathrm{e}}^{-1.2\mathrm{t}}\phantom{\rule{0ex}{0ex}}$

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

$12-1.2\mathrm{v}=6.{\mathrm{e}}^{-1.2\mathrm{t}}\phantom{\rule{0ex}{0ex}}\mathrm{v}\left(\mathrm{t}\right)=10-5.{\mathrm{e}}^{-1.2\mathrm{t}}\phantom{\rule{0ex}{0ex}}$

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

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

The limiting velocity of the sailboat is

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

**Hence, the Limiting velocity is 10m/sec.**

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