/*! 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 14 Galileo developed a formula for ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 14

Galileo developed a formula for a body's velocity during free fall by rolling balls from rest down increasingly steep inclined planks and looking for a limiting formula that would predict a ball's behavior when the plank was vertical and the ball fell freely; see part (a) of the accompanying figure. He found that, for any given angle of the plank, the ball's velocity \(t\) seconds into motion was a constant multiple of \(t .\) That is, the velocity was given by a formula of the form \(v=k t .\) The value of the constant \(k\) depended on the inclination of the plank. In modern notation-part (b) of the figure-with distance in meters and time in seconds, what Galileo determined by experiment was that, for any given angle \(\theta,\) the ball's velocity \(t\) s into the roll was $$v=9.8(\sin \theta) t \mathrm{m} / \mathrm{s}.$$ a. What is the equation for the ball's velocity during free fall? b. Building on your work in part (a), what constant acceleration does a freely falling body experience near the surface of Earth?

Short Answer

Expert verified
a. The velocity equation is \( v = 9.8t \). b. The constant acceleration is 9.8 m/s虏.

Step by step solution

01

Understanding the Problem

We are given that the velocity of a ball rolling down an inclined plane is given by the formula \( v = 9.8 \sin(\theta) t \). For free fall, the plank is vertical, which means the angle \( \theta = 90^\circ \). Our task is to find the velocity equation during free fall and then determine the constant acceleration.
02

Calculate the Velocity for Free Fall

When the plank is vertical, \( \theta = 90^\circ \). The sine of 90 degrees is 1. Thus, the equation for velocity during free fall becomes: \[ v = 9.8 \sin(90^\circ) t = 9.8 \cdot 1 \cdot t = 9.8t \].
03

Determine the Constant Acceleration

Acceleration is the rate of change of velocity. For a freely falling object, the velocity formula is \( v = 9.8t \). Since acceleration is defined as \( a = \frac{dv}{dt} \), differentiating \( v = 9.8t \) with respect to \( t \) gives \( a = 9.8 \). This is the constant acceleration.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Velocity in Free Fall
In physics, free fall refers to the movement of an object under the influence of gravitational force only. This means there are no other forces acting, such as air resistance. Now, let's look into how Galileo's experiments give us a clear understanding of this concept.

Galileo discovered that when an object is in free fall, its velocity increases constantly over time. This is because the object accelerates due to gravity. A key finding from his experiments is the formula that gives the object's velocity in free fall:
  • Velocity: \( v = 9.8t \)
  • Where \( v \) is the velocity in meters per second (m/s)
  • \( t \) is the time in seconds
This equation shows that velocity is directly proportional to time. As time increases, the velocity of the object also increases at a constant rate.
By using this equation, we can predict how fast an object will be moving at any point during its free fall, provided we know how long it has been falling. This is crucial for understanding the dynamics of motion under gravity and forms the basis for many principles in mechanics.
Inclined Plane Experiments
Galileo's inclined plane experiments were pioneering efforts to study the acceleration of objects. Before he studied objects in free fall, he rolled spheres down inclined planes. This allowed him to carefully observe how objects accelerate under a constant force with reduced friction and more controlled conditions.

The setup was ingenious because it slowed down the motion, letting Galileo and his students measure time and distance more accurately. When balls rolled down different angles of incline, he observed that regardless of the angle, the velocity at any given time was a constant multiple of time. This is expressed in the formula:
  • Velocity on an inclined plane: \( v = 9.8 \sin(\theta) t \)
  • \(\theta\) is the angle of the incline
  • If the slope is vertical, \(\theta = 90^\circ \) & \(\sin(90^\circ) = 1\)
By gradually increasing the plane's steepness, he noted that when the angle reached 90 degrees, the motion mimicked free fall perfectly. This led to the modern understanding of free fall acceleration on Earth being akin to dropping from vertical height.
Constant Acceleration Due to Gravity
The concept of constant acceleration is pivotal in understanding how objects behave when they are in free fall. Near the surface of Earth, objects experience a constant acceleration due to gravity (denoted as \( g \)).

Galileo鈥檚 experiments showed that this acceleration equals approximately 9.8 meters per second squared (m/s虏). This number comes from the equation of motion for free fall, where the velocity increases linearly with time:
  • \( a = \frac{dv}{dt} \)
  • For \( v = 9.8t \), that derivative gives \( a = 9.8 \)
This indicates that if you drop an object, its speed will increase by roughly 9.8 m/s for every second it falls, assuming no air resistance.

This constant acceleration is fundamental for calculations related to motion and is used extensively in physics to solve problems involving falling objects. It's a universal figure that applies irrespective of the object's mass, which means that in a vacuum, all objects fall at the same rate. This principle revolutionized how we understand motion and set the foundation for Newton's Laws of Motion.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval \(I\). Perform the following steps: a. Plot the function \(f\) over \(I\) b. Find the linearization \(L\) of the function at the point \(a\) c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta>0\) as you can, satisfying \(|x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon\) for \(\epsilon=0.5,0.1,\) and \(0.01 .\) Then check graphically to see if your \(\delta\) -estimate holds true. $$f(x)=\frac{x-1}{4 x^{2}+1}, \quad\left[-\frac{3}{4}, 1\right], \quad a=\frac{1}{2}$$

Give the position function \(s=f(t)\) of an object moving along the \(s\) -axis as a function of time \(t\). Graph \(f\) together with the velocity function \(v(t)=d s / d t=f^{\prime}(t)\) and the acceleration function \(a(t)=d^{2} s / d t^{2}=f^{\prime \prime}(t) .\) Comment on the object's behavior in relation to the signs and values of \(v\) and \(a .\) Include in your commentary such topics as the following: a. When is the object momentarily at rest? b. When does it move to the left (down) or to the right (up)? c. When does it change direction? d. When does it speed up and slow down? e. When is it moving fastest (highest speed)? Slowest? f. When is it farthest from the axis origin? $$s=4-7 t+6 t^{2}-t^{3}, \quad 0 \leq t \leq 4$$

Suppose that \(y=f(x)\) is differentiable at \(x=a\) and that \(g(x)=\) \(m(x-a)+c\) is a linear function in which \(m\) and \(c\) are constants. If the error \(E(x)=f(x)-g(x)\) were small enough near \(x=a\) we might think of using \(g\) as a linear approximation of \(f\) instead of the linearization \(L(x)=f(a)+f^{\prime}(a)(x-a) .\) Show that if we impose on \(g\) the conditions 1\. \(E(a)=0\) 2\. \(\lim _{x \rightarrow a} \frac{E(x)}{x-a}=0\) then \(g(x)=f(a)+f^{\prime}(a)(x-a) .\) Thus, the linearization \(L(x)\) gives the only linear approximation whose error is both zero at \(x=a\) and negligible in comparison with \(x-a\)

In the late 1860 s, Adolf Fick, a professor of physiology in the Faculty of Medicine in W眉rzberg, Germany, developed one of the methods we use today for measuring how much blood your heart pumps in a minute. Your cardiac output as you read this sentence is probably about \(7 \mathrm{L} / \mathrm{min} .\) At rest it is likely to be a bit under \(6 \mathrm{L} / \mathrm{min}\). If you are a trained marathon runner running a marathon, your cardiac output can be as high as \(30 \mathrm{L} / \mathrm{min}\). Your cardiac output can be calculated with the formula $$y=\frac{Q}{D},$$ where \(Q\) is the number of milliliters of \(\mathrm{CO}_{2}\) you exhale in a minute and \(D\) is the difference between the \(\mathrm{CO}_{2}\) concentration \((\mathrm{mL} / \mathrm{L})\) in the blood pumped to the lungs and the \(\mathrm{CO}_{2}\) concentration in the blood returning from the lungs. With \(Q=233 \mathrm{mL} / \mathrm{min}\) and \(D=97-56=41 \mathrm{mL} / \mathrm{L}\), \(y=\frac{233 \mathrm{mL} / \mathrm{min}}{41 \mathrm{mL} / \mathrm{L}} \approx 5.68 \mathrm{L} / \mathrm{min}\), fairly close to the \(6 \mathrm{L} / \mathrm{min}\) that most people have at basal (resting) conditions. (Data courtesy of J. Kenneth Herd, M.D., Quillan College of Medicine, East Tennessee State University.) Suppose that when \(Q=233\) and \(D=41,\) we also know that \(D\) is decreasing at the rate of 2 units a minute but that \(Q\) remains unchanged. What is happening to the cardiac output?

What happens if you can write a function as a composite in different ways? Do you get the same derivative each time? The Chain Rule says you should. Try it with the functions. Find \(d y / d x\) if \(y=x^{3 / 2}\) by using the Chain Rule with \(y\) as a composite of a. \(y=u^{3}\) and \(u=\sqrt{x}\) b. \(y=\sqrt{u}\) and \(u=x^{3}\)
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.