/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 A motorboat weighs \(32,000 \mat... [FREE SOLUTION] | 91Ó°ÊÓ

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A motorboat weighs \(32,000 \mathrm{lb}\) and its motor provides \(\mathrm{a}\) thrust of 5000 lb. Assume that the water resistance is 100 pounds for each foot per second of the speed \(v\) of the boat. Then $$ 1000 \frac{d v}{d t}=5000-100 v $$ If the boat starts from rest, what is the maximum velocity that it can attain?

Short Answer

Expert verified
The maximum velocity the boat can attain is 50 ft/s.

Step by step solution

01

Understanding the Differential Equation

First, we are given the differential equation \(1000 \frac{dv}{dt} = 5000 - 100v\). This equation models the change in speed of the motorboat over time. The term \(5000\) represents the constant thrust from the motor, while \(100v\) represents the water resistance which is proportional to the velocity. Our task is to find the maximum steady-state (asymptotic) velocity the boat can achieve.
02

Setting Up the Steady-State Equation

To find the maximum velocity, we assume the velocity \(v\) becomes constant over time. This means \(\frac{dv}{dt} = 0\). Substituting \(0\) for \(\frac{dv}{dt}\) in the differential equation, we get: \(0 = 5000 - 100v\). This tells us the point at which the thrust and resistance forces are balanced, and velocity stops changing.
03

Solving for Maximum Velocity

Rearrange the equation \(0 = 5000 - 100v\) to solve for \(v\). Add \(100v\) to both sides to get \(100v = 5000\). Then, divide both sides by 100: \(v = \frac{5000}{100} = 50\). Thus, the maximum velocity that the motorboat can attain is \(50\) feet per second.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Steady-State Velocity
In the world of differential equations and dynamics, a term that often comes up is "steady-state velocity". This is essentially the velocity at which an object stops accelerating and moves at a constant speed. When something reaches a steady-state, it means the forces acting upon it are perfectly balanced. In simple terms, the push equals the pull.

In our motorboat example, the steady-state velocity is reached when the thrust from the motor is exactly canceled out by the water resistance. This balance is where the change in velocity (\(\frac{dv}{dt}\)) becomes zero.

- A steady-state does not mean the object isn't moving; it just means its speed isn't increasing or decreasing.- For the motorboat, the steady-state velocity is identified mathematically as a point where thrust (5000 lbs force) and resistance (100 lbs for every foot/second of speed) are equal.

This principle can be applied in many real-world situations, from cars on a highway to airplanes flying at cruise altitude.
Water Resistance
Water resistance, or drag, is a force that opposes the motion of objects through water. In physics, it is a type of friction that acts against the forward motion. It's crucial to understand how water resistance affects a motorboat's movement.

The water resistance the motorboat faces in our exercise is directly proportional to its speed. Each increase in speed by 1 foot per second results in an additional 100 pounds of resistance. This has the effect of slowing down the boat as it speeds up, reaching a point where the increase in speed results in a proportionate increase in resistance.

Key Points about Water Resistance:- It depends on the speed and shape of the object moving through the fluid.- As speed increases, so does water resistance, typically requiring more power to maintain or increase speed further.- Understanding its influence is critical for calculating maximum achievable velocities in fluid dynamics scenarios.

This relationship is captured by our equation: \(100v\), indicating any motorboat's speed and encountering more resistance with every additional foot per second, highlighting the direct impact of drag on motorboat dynamics.
Motorboat Dynamics
The dynamics of a motorboat combine various forces that dictate how it moves. To navigate efficiently, understanding the input (the motor's thrust) and the opposition (water resistance) is essential.

In our equation,\(1000 \frac{dv}{dt} = 5000 - 100v\), we see a clear relationship between the thrust generated by the motorboat's engine and the opposing drag due to water. Here is how these dynamics unfold:

- **Thrust**: This is the driving force produced by the motor, stated as 5000 pounds. It propels the boat forward, counteracting water resistance.- **Water Resistance**: As previously mentioned, this increases with speed, acting as the primary opposing force.

The equation essentially tells us that the equalization of these forces indicates where the boat's speed will no longer change — this is when it has maxed out its potential under the given conditions. Understanding these factors can help optimize boat design and engine efficiency, ensuring maximum performance on water.

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Most popular questions from this chapter

A motorboat starts from rest (initial velocity \(v(0)=v_{0}=\) 0). Its motor provides a constant acceleration of \(4 \mathrm{ft} / \mathrm{s}^{2}\) but water resistance causes a deceleration of \(v^{2} / 400 \mathrm{ft} / \mathrm{s}^{2}\) Find \(v\) when \(t=10 \mathrm{~s}\), and also find the limiting velocity as \(t \rightarrow+\infty\) (that is, the maximum possible speed of the boat).

A computer with a printer is required. In these initial value problems, use the improved Euler method with step sizes \(h=0.1,0.02,0.004\), and \(0.0008\) to approximate to five decimal places the values of the solution at ten equally spaced points of the given interval. Print the results in tabular form with appropriate headings to make it easy to gauge the effect of varying the step size h. Throughout, primes denote derivatives with respect to \(x\). $$ y^{\prime}=\frac{x}{1+y^{2}}, y(-1)=1 ;-1 \leqq x \leqq 1 $$

Suppose that a body moves through a resisting medium with resistance proportional to its velocity \(v\), so that \(d v / d t=-k v .\) (a) Show that its velocity and position at time \(t\) are given by $$ v(t)=v_{0} e^{-k t} $$ and $$ x(t)=x_{0}+\left(\frac{v_{0}}{k}\right)\left(1-e^{-k t}\right) $$ (b) Conclude that the body travels only a finite distance, and find that distance.

Suppose that a projectile is fired straight upward from the surface of the earth with initial velocity \(v_{0}<\sqrt{2 G M / R}\) Then its height \(y(t)\) above the surface satisfies the initial value problem $$ \frac{d^{2} y}{d t^{2}}=-\frac{G M}{(y+R)^{2}} ; \quad y(0)=0, \quad y^{\prime}(0)=v_{0} $$ Substitute \(d v / d t=v(d v / d y)\) and then integrate to obtain $$ v^{2}=v_{0}^{2}-\frac{2 G M y}{R(R+y)} $$ for the velocity \(v\) of the projectile at height \(y\). What maximum altitude does it reach if its initial velocity is \(1 \mathrm{~km} / \mathrm{s}\) ?

Use Euler's method with a computer system to find the desired solution values. Start with step size \(h=0.1\), and then use successively smaller step sizes until successive approximate solution values at \(x=2\) agree rounded off to two decimal places. \(y^{\prime}=x+\frac{1}{2} y^{2}, y(-2)=0 ; y(2)=?\)

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