Chapter 3: Q3.4-22E (page 117)

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

Expert verified

The limiting velocity of the sailboat is 10 m/second.
The equation of motion is $x (t) = 10 t + \frac{25}{6} (e^{- 1.2 t} - 1)$ .

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.

$F = 600 N F_{2} = 60 v$

Now

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

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

$12 - 1.2 v = 6 . e^{- 1.2 t} v (t) = 10 - 5 . e^{- 1.2 t}$

Find the value of equation of motion

$x (t) = \int_{0}^{t} v (s) ds x (t) = \int_{0}^{t} 10 - 5 e^{- 1.2 s} ds = {[10 s + \frac{5 e^{- 1.2 s}}{1.2}]}_{0}^{t} x (t) = 10 t + \frac{25}{6} (e^{- 1.2 t} - 1)$

Hence,The equation of motion is $x (t) = 10 t + \frac{25}{6} (e^{- 1.2 t} - 1)$ .

Find the limiting velocity

The limiting velocity of the sailboat is

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

Hence, the Limiting velocity is 10m/sec.

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Find the equation of velocity

Find the value of equation of motion

Find the limiting velocity

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