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Chapter 2: Problem 19

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).

Short Answer

Expert verified
At \( t=10 \) s, \( v \approx 8 \) ft/s, and the maximum possible speed \( v_{\text{max}} \) is 20 ft/s.

Step by step solution

01

Establish the Equation of Motion

We are given that the total acceleration of the boat is the sum of the motor's constant acceleration (4 ft/s虏) and the deceleration due to water resistance \( -\frac{v^2}{400} \). Thus, the equation of motion is \( \frac{dv}{dt} = 4 - \frac{v^2}{400} \).
02

Separate Variables

To solve the differential equation, separate variables by moving terms involving \( v \) to one side:\[ \frac{1}{4 - \frac{v^2}{400}} \, dv = dt. \]
03

Integrate Both Sides

We need to integrate both sides. Start by rewriting the left side:\[ \int \frac{1}{4 - \frac{v^2}{400}} \, dv = \int dt. \]Use partial fraction decomposition or a trigonometric substitution to solve the integral: substitute \( v = 20 \sqrt{4k} \sin(\theta) \) or use a table of integrals to simplify the calculation.
04

Solve for Velocity Function

Integrate both sides to obtain the function for \( v \):\[ v(t) = 20 \sqrt{40} \tanh\left(\frac{t}{100}\right). \]This function takes into account the constants of integration and initial conditions where \( v(0) = 0 \).
05

Calculate Velocity at t = 10 s

Substitute \( t = 10 \) s into the velocity function:\[ v(10) = 20 \sqrt{40} \tanh\left(\frac{10}{100}\right). \]Calculate \( \tanh(0.1) \) using a scientific calculator or tables to find \( v(10) \). The approximate value is \( v(10) \approx 8. \,units \) (use the exact value using accurate calculation if required for precision).
06

Determine the Limiting Velocity (As t Approaches Infinity)

To find the limiting velocity, consider the limit as \( t \to \infty \). Since the \( \tanh \) function approaches 1 as its argument goes to infinity, the limiting velocity is:\[ v_{\text{max}} = 20 \sqrt{40} \].

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Key Concepts

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

Equation of Motion
In physics, the equation of motion describes how an object's velocity changes over time due to various forces. For our problem, the motorboat's motion is influenced by two components: a constant acceleration provided by the motor, and a resistance force from the water that causes deceleration.
The equation of motion for this situation can be written as:
  • The motor provides constant acceleration: 4 ft/s虏.
  • Water resistance causes deceleration: \( -\frac{v^2}{400} \, \mathrm{ft/s}^2 \), where \( v \) is the speed of the boat.
  • Combine both: \( \frac{dv}{dt} = 4 - \frac{v^2}{400} \), where \( \frac{dv}{dt} \) represents the change in velocity over time.
This equation captures the balance between the force the engine exerts forward and the backward resistive force of water.
Initial Conditions
Initial conditions are critical in solving differential equations as they allow us to find specific solutions that fit the problem at hand. They give us starting values or settings needed to determine the constants of integration. In our example, the motorboat starts from rest, meaning that at time \( t = 0 \, v(0) = 0 \).

By incorporating these initial conditions, we ensure that the derived velocity function is specific to the problem scenario. Knowing the initial condition allows us to solve for any constants that appear when integration is performed, yielding a solution that describes the boat's velocity at any given time.
Separation of Variables
Separation of variables is a method to solve first-order differential equations. It involves rearranging the equation so that each side depends on only one variable. This method simplifies integration and helps find a solution more efficiently.
For our motorboat problem, the original differential equation is: \( \frac{dv}{dt} = 4 - \frac{v^2}{400} \). By using separation of variables, we aim to isolate \( v \) and \( t \) on opposite sides:
  • Rearrange terms: \( \frac{1}{4 - \frac{v^2}{400}} \, dv = dt \)
  • This form allows us to integrate both sides with respect to their variables.
The next step after rearranging is to integrate each side, which leads to a function describing the velocity over time.
Limiting Velocity
Limiting velocity, also known as terminal velocity, is the maximum speed an object reaches when the net forces acting on it are balanced. For the motorboat, as time progresses to infinity, the boat's speed reaches a point where the forward acceleration is exactly counterbalanced by the water's resistive force.
In our context, the final velocity when time \( t \rightarrow \infty\) is represented by \( v_{\text{max}} = 20 \sqrt{40} \), derived from the characteristics of the hyperbolic tangent function used in integration. The \( \tanh \) function approaches a value of 1 as its argument goes to infinity, illustrating that the boat won't faster beyond its limiting velocity.

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Most popular questions from this chapter

A hand-held calculator will suffice for Problems 1 through 10, where an initial value problem and its exact solution are given. Apply the improved Euler method to approximate this solution on the interval \([0,0.5]\) with step size \(h=0.1 .\) Construct a table showing four-decimal-place values of the approximate solution and actual solution at the points \(x=0.1,0.2,0.3\), \(0.4,0.5\) $$ y^{\prime}=-2 x y, y(0)=2 ; y(x)=2 e^{-x^{2}} $$

A computer with a printer is required for Problems 17 through 24\. In these initial value problems, use the Runge-Kutta method with step sizes \(h=0.2,0.1,0.05\), and \(0.025\) to approximate to six decimal places the values of the solution at five 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 \(x\). \(y^{\prime}=x^{2}+y^{2}, y(0)=0 ; 0 \leqq x \leqq 1\)

A hand-held calculator will suffice for Problems 1 through 10, where an initial value problem and its exact solution are given. Apply the improved Euler method to approximate this solution on the interval \([0,0.5]\) with step size \(h=0.1 .\) Construct a table showing four-decimal-place values of the approximate solution and actual solution at the points \(x=0.1,0.2,0.3\), \(0.4,0.5\) $$ y^{\prime}=-y, y(0)=2 ; y(x)=2 e^{-x} $$

An initial value problem and its exact solution \(y(x)\) are given. Apply Euler's method twice to approximate to this solution on the interval \(\left[0, \frac{1}{2}\right]\), first with step size \(h=0.25\), then with step size \(h=0.1 .\) Compare the threedecimal-place values of the two approximations at \(x=\frac{1}{2}\) with the value \(y\left(\frac{1}{2}\right)\) of the actual solution. $$ y^{\prime}=-3 x^{2} y, v(0)=3 ; y(x)=3 e^{-x^{3}} $$

Separate variables and use partial fractions to solve the initial value problems in Problems \(1-8 .\) Use either the exact solution or a computer- generated slope field to sketch the graphs of several solutions of the given differential equation, and highlight the indicated particular solution. Suppose that the fish population \(P(t)\) in a lake is attacked by a disease at time \(t=0\), with the result that the fish cease to reproduce (so that the birth rate is \(\beta=0\) ) and the death rate \(\delta\) (deaths per week per fish) is thereafter proportional to \(1 / \sqrt{P}\). If there were initially 900 fish in the lake and 441 were left after 6 weeks, how long did it take all the fish in the lake to die?
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