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Consider an object on which the net force is a resistive force proportional to the square of its speed. For example, assume that the resistive force acting on a speed skater is \(f=-k m v^{2},\) where \(k\) is a constant and \(m\) is the skater's mass. The skater crosses the finish line of a straight-line race with speed \(v_{0}\) and then slows down by coasting on his skates. Show that the skater's speed at any time \(t\) after crossing the finish line is \(v(t)=v_{0} /\left(1+k t v_{0}\right) .\) This problem also provides the background for the two following problems.

Short Answer

Expert verified
The skater's speed at any time \(t\) after crossing the finish line is given by the function \(v(t)=\frac{v_0}{1+k t v_0}\).

Step by step solution

01

Understand the problem

We need to find an expression for the skater's velocity as a function of time, given that the force acting on the skater is resistive and proportional to the square of the skater's velocity. We are given that the resistive force is represented by the equation \( f = -k m v^2 \), where \( k \) is a constant, \( m \) is the skater's mass, and the skater's initial velocity as he crosses the finish line is \( v_0 \).
02

Write down Newton's second law

According to Newton's second law, the net force acting on an object is equal to the mass times the acceleration of the object (\( F = m a \)). Since the only force acting on the skater is the resistive force, we can equate \( m a \) with the resistive force: \( m a = -k m v^2 \).
03

Simplify the equation

The mass \( m \) is on both sides of the equation, so it cancels out. This simplification yields the differential equation for acceleration (which is the derivative of velocity with respect to time): \( a = \frac{dv}{dt} = -k v^2 \).
04

Separate variables

To solve the differential equation, separate the variables (velocity and time): \( \frac{dv}{v^2} = -k dt \).
05

Integrate both sides

Integrate the left side with respect to velocity and the right side with respect to time: \( \int \frac{dv}{v^2} = -\int k dt \). The integral on the left side is \( -\frac{1}{v} \) (up to a constant), and the integral on the right side is \( -kt \) (also up to a constant).
06

Solve the integral equation

The result of integration is \( -\frac{1}{v} = -kt + C \), where \( C \) is the integration constant. This equation can also be written as \( \frac{1}{v} = kt - C \).
07

Determine the constant of integration C

To find \( C \), use the initial condition that when \( t = 0 \), the velocity \( v = v_0 \). Substituting these values into the equation gives \( \frac{1}{v_0} = -C \), so the constant \( C \) is equal to \( -\frac{1}{v_0} \). The equation for velocity as a function of time is now \( \frac{1}{v} = kt + \frac{1}{v_0} \).
08

Express v as a function of time

Finally, solve for \( v(t) \) by taking the reciprocal of both sides and simplifying: \( v(t) = \frac{1}{kt + \frac{1}{v_0}} = \frac{v_0}{1 + k t v_0} \). This is the skater's speed at any time \( t \) after crossing the finish line.

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Key Concepts

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

Resistive Force
Resistive forces play a key role in various physical phenomena, especially in the context of motion. They are forces that oppose the motion of an object through a medium, such as air or water. In our exercise, the resistive force is proportional to the square of the speed skater's velocity, exemplified by the formula \(f = -k m v^2\). The \(-\) sign indicates that this force acts in the opposite direction to the movement.

Understanding this force is critical for predicting how objects will slow down over time when no other forces are applied. For instance, our speed skater experiences this force after crossing the finish line and consequently decelerates. In real-life situations, this could relate to vehicles like cars and bicycles slowing down when the engine or pedals are no longer powering them, or to skydivers falling through the air at terminal velocity. Mathematically, this relationship is crucial for setting up differential equations that describe the motion of objects under such conditions.

Application in Physics

Resistive forces, particularly those that are velocity-dependent, require careful consideration when designing objects that move through fluids, like ships and airplanes, or when calculating the landing trajectories of space exploration vehicles. By understanding the dynamics of resistive forces, engineers can optimize designs to minimize energy losses and enhance performance.
Newton's Second Law
Newton's second law is at the heart of classical mechanics and is expressed as \( F = m a \), where \( F \) stands for the net force applied to an object, \( m \) is the object's mass, and \( a \) is the acceleration. This fundamental principle states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.

In the world of physics, this law is the framework for analyzing how forces affect the motion of objects. It allows us to predict the resulting acceleration for an object subjected to any given set of forces. In our exercise, we applied Newton's second law to the motion of the speed skater. As the skater is subject only to the resistive force and no additional forces, the equation simplifies to \( m a = -k m v^2 \). This equation forms the basis for deriving the differential equation that we need to solve to find the skater's velocity as a function of time.

Real-World Importance

Newton's second law is not only essential for theoretical exercises but also for practical applications. For example, engineers use it to calculate the forces required to move elevators, while car manufacturers rely on it to design safe and efficient braking systems. It is a versatile tool that can be applied across various aspects of physics and engineering to understand and manipulate the forces of nature.
Velocity-Time Relationship
The velocity-time relationship describes how the velocity of an object changes with time. In the context of differential equations and physics, we're often interested in cases where this relationship is not constant, as is the case with our speed skater example. The derived formula \( v(t) = \frac{v_0}{1 + k t v_0} \) shows how the skater's velocity decreases over time due to the resistive force.

This equation is a solution to a differential equation and it represents a model of deceleration where the velocity is inversely related to time. From this expression, it's clear that as time \( t \) increases, the velocity \( v(t) \) decreases. Such velocity-time relationships are fundamental to understanding motion, predicting future positions, and determining how long it takes for an object to stop moving under certain conditions.

Broader Implications

The velocity-time relationship is not only about plotting speed against time in physics classes. It's a crucial concept for designing timing mechanisms in sports, programming the motion of robots or animations in video games, and even controlling the speed of chemical reactions in dynamic systems. This relationship forms the basis of kinematics, which is the study of motion without considering the forces that cause it, and plays a tremendous role in transportation, sports science, and any technology involving movement.

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Most popular questions from this chapter

A light string can support a stationary hanging load of 25.0 \(\mathrm{kg}\) before breaking. A 3.00 -kg object attached to the string rotates on a horizontal, frictionless table in a circle of radius \(0.800 \mathrm{m},\) while the other end of the string is held fixed. What range of speeds can the object have before the string breaks?

The pilot of an airplane executes a constant-speed loop-the-loop maneuver in a vertical circle. The speed of the airplane is 300 \(\mathrm{mi} / \mathrm{h}\) and the radius of the circle is 1200 \(\mathrm{ft}\) . (a) What is the pilot's apparent weight at the lowest point if his true weight is 160 \(\mathrm{lb}\) ? (b) What is his apparent weight at the highest point? (c) What If? Describe how the pilot could experience weightlessness if both the radius and the speed can be varied. (Note: His apparent weight is equal to the magnitude of the force exerted by the seat on his body.)

The cornering performance of an automobile is evaluated on a skidpad, where the maximum speed that a car can maintain around a circular path on a dry, flat surface is measured. Then the centripetal acceleration, also called the lateral acceleration, is calculated as a multiple of the free-fall acceleration \(g\) . The main factors affecting the per system of the car. A Dodge Viper GTS can negotiate a skid- pad of radius 61.0 \(\mathrm{m}\) at 86.5 \(\mathrm{km} / \mathrm{h}\) . Calculate its maximum lateral acceleration.

(a) A luggage carousel at an airport has the form of a section of a large cone, steadily rotating about its vertical axis. Its metallic surface slopes downward toward the outside, making an angle of \(20.0^{\circ}\) with the horizontal. A piece of luggage having mass 30.0 \(\mathrm{kg}\) is placed on the carousel, 7.46 \(\mathrm{m}\) from the axis of rotation. The travel bag goes around once in 38.0 s. Calculate the force of static friction between the bag and the carousel. (b) The drive motor is shifted to turn the carousel at a higher constant rate of rotation, and the piece of luggage is bumped to another position, 7.94 \(\mathrm{m}\) from the axis of rotation. Now going around once in every 34.0 \(\mathrm{s}\) , the bag is on the verge of slipping. Calculate the coefficient of static friction between the bag and the carousel.

A student stands in an elevator that is continuously accelerating upward with acceleration \(a\) . Her backpack is sitting on the floor next to the wall. The width of the elevator car is \(L\) . The student gives her backpack a quick kick at \(t=0\) , imparting to it speed \(v\) , and making it slide across the elevator floor. At time \(t\) , the backpack hits the opposite wall. Find the coefficient of kinetic friction \(\mu_{k}\) between the backpack and the elevator floor.

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