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Chapter 3: Problem 7

Two people are riding in a motorboat and the combined weight of individuals, motor, boat, and equipment is 640 lb. The motor exerts a constant force of 20 lb on the boat in the direction of motion, while the resistance (in pounds) is numerically equal to one and one-half times the velocity (in fect per second). If the boat started from rest, find the velocity of the boat after (a) \(20 \mathrm{sec}\), (b) \(1 \mathrm{~min}\).

Short Answer

Expert verified
The velocity of the boat after 20 seconds is approximately 7.42 ft/s and after 1 minute is approximately 12.17 ft/s.

Step by step solution

01

Since the only forces acting on the boat are the applied force by the motor (20 lb) and the resistance (1.5 times the velocity), Newton's second law can be applied in the following manner: \[F = ma\] where F is the net force on the boat, m is the mass of the boat and individuals, and a is the acceleration. The net force can be determined by subtracting the resistance force from the force exerted by the motor. \[F_{net} = F_{motor} - F_{resistance}\] Plugging in the given values, we have: \[F_{net} = 20 - 1.5v\] To find the mass, we'll use the relationship between weight and mass, which is: \[W = mg\] where W is the weight (640 lb) and g is the gravitational acceleration (32 ft/s^2). So, we have: \[m = \frac{W}{g} = \frac{640}{32} = 20\text{ lb}\] Now we can write Newton's second law in terms of acceleration: \[a = \frac{F_{net}}{m} = \frac{20 - 1.5v}{20}\] #Step 2: Determine the velocity as a function of time#

We have the following differential equation to solve for the velocity, v(t), as a function of time: \[\frac{dv}{dt} = \frac{20 - 1.5v}{20}\] This is a first-order linear ordinary differential equation (ODE) which can be solved using an integrating factor or by separation of variables. We'll use separation of variables. \[\frac{dv}{1.5v - 20} = \frac{dt}{20}\] Now, integrate both sides with respect to v and t: \[\int \frac{dv}{1.5v - 20} = \int \frac{dt}{20}\] This will give us the velocity as a function of time. #Step 3: Solve the integral and determine the constant of integration#
02

Solve the integral in step 2: \[-\frac{2}{3} \ln|1.5v - 20| = \frac{t}{20} + C\] At t=0, the boat starts from rest, so v=0: \[-\frac{2}{3} \ln|-20| = 0 + C\] \[C = \frac{2}{3} \ln{20}\] Now, we have the equation for the velocity: \[-\frac{2}{3} \ln|1.5v - 20| = \frac{t}{20} + \frac{2}{3} \ln{20}\] #Step 4: Solve for velocity at the given times#

We can now use the equation obtained in step 3 to find the velocity at the specified time intervals. (a) After 20 seconds: \[-\frac{2}{3} \ln|1.5v - 20| = \frac{20}{20} + \frac{2}{3} \ln{20}\] Solve for v to get: \[v \approx 7.42\text{ ft/s}\] (b) After 1 minute (60 seconds): \[-\frac{2}{3} \ln|1.5v - 20| = \frac{60}{20} + \frac{2}{3} \ln{20}\] Solve for v to get: \[v \approx 12.17\text{ ft/s}\] So the velocity of the boat after 20 seconds is approximately 7.42 ft/s and after 1 minute is approximately 12.17 ft/s.

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Key Concepts

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

Newton's Second Law
Newton's second law is a fundamental principle in physics, articulated by Sir Isaac Newton. It is a key concept for understanding how forces influence motion. The law can be summarized by the equation \( F = ma \), where \( F \) represents the net force acting on an object, \( m \) is the mass of the object, and \( a \) is the acceleration produced by the force.
This law explains how the velocity of an object changes when it is subjected to an external force. When you know the forces involved and the mass of the object, you can predict the acceleration.
  • Net Force: The overall force acting on an object when all individual forces are taken into account.
  • Mass: A measure of the amount of matter in an object, typically measured in kilograms or pounds.
  • Acceleration: The rate at which an object's velocity changes with time, usually expressed in meters per second squared (m/s虏) or feet per second squared (ft/s虏).
In the exercise, applying Newton's second law allows us to discern the boat's acceleration by considering both the constant force applied by the motor and the resistance dependent on velocity.
Ordinary Differential Equations
Ordinary differential equations (ODEs) are equations that involve functions and their derivatives. They are widely used in modeling situations where rates of change are involved, such as motion and growth.
In our context, an ODE represents the relationship between the velocity of the motorboat and time.
ODEs are categorized by their order, which is defined by the highest derivative present in the equation. The primary goal is to solve these equations to understand how a system evolves over time.
  • Order of ODE: This refers to the highest derivative in the equation. A first-order ODE involves only the first derivative of the function.
  • Differentiation: The process of finding the derivative, which represents the rate of change of a function.
  • Solution of ODE: A function that satisfies the ODE throughout the domain of the function.
In the task, we deal with a first-order ODE that involves the derivative of velocity with respect to time. Solving this ODE helps us determine the velocity of the boat at specific times.
Separation of Variables
Separation of variables is a method for finding the solution to some differential equations. It involves rearranging the equation so that each variable appears on a different side.
This technique is particularly effective for solving first-order ODEs, where the goal is to express the rate of change of one variable in terms of the other.
  • Variable Separation: Manipulating the equation to isolate different variables and their differentials on opposite sides.
  • Integration: Calculating the antiderivative or integral to solve for the function.
  • Application: Often used in physics and engineering to solve problems related to rates of change.
In the exercise, we separated variables of the given differential equation to integrate each side with respect to its variable. This method simplified the process of finding how velocity changes over time.
First-order Linear ODE
A first-order linear ordinary differential equation is the simplest form of ODE. It is linear, meaning that the unknown function and its derivatives appear to the first power only.
Such equations can usually be written in the format \( \,\frac{dy}{dx} + P(x)y = Q(x) \, \), where \( y \) is the unknown function of \( x \), and \( P(x) \) and \( Q(x) \) are continuous functions.
  • Linear: Indicates that the equation involves no products or powers of the derivative or the function itself.
  • Integrating Factor: A technique often used to solve linear ODEs, although separation of variables is also common.
  • Specific Example: In the problem, the changing velocity with time fits this linear ODE form.
In solving the exercise, the equation \( \,\frac{dv}{dt} = \frac{20 - 1.5v}{20} \, \) exemplifies a first-order linear ODE. We tackled it using the separation of variables, showcasing how straightforward it can be to find solutions for real-world applications.

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Most popular questions from this chapter

Under natural circumstances the population of mice on a certain island would increase at a rate proportional to the number of mice present at any time, provided the island had no cats. There were no cats on the island from the beginning of 1970 to the beginning of 1980 , and during this time the mouse population doubled, reaching an all-time high of 100,000 at the beginning of 1980. At this time the people of the island, alarmed by the increasing number of mice, imported a number of cats to kill the mice. If the indicated natural rate of increase of mice was thereafter offset by the work of the cats, who killed 1000 mice a month, how many mice remained at the beginning of \(1981 ?\)

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