/*! 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 104 Determine whether the statement ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 104

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If the velocity of an object is constant, then its acceleration is zero.

Short Answer

Expert verified
The statement 'If the velocity of an object is constant, then its acceleration is zero' is true. Because if an object's velocity is constant (i.e., its speed and direction doesn't change), there is no change in velocity, therefore, the acceleration, which is the rate of change of velocity, is zero.

Step by step solution

01

Understand the definitions of Velocity and Acceleration

Velocity is a quantity that describes the speed and direction of an object's motion. Acceleration, meanwhile, is the rate of change of an object's velocity per unit time.
02

Analyze the statement

The given statement says that if the velocity of any object is constant, then the object's acceleration is zero. We know that velocity is speed in a particular direction, and a constant velocity implies that the speed and direction remain unchanged over time. Acceleration is the rate of change of velocity - it's how much the velocity changes per unit time. Now, if velocity is constant, that means the velocity is not changing over time. Therefore, since acceleration is the rate of change of velocity and the velocity isn't changing, then the acceleration is zero.
03

Conclusion

Based on the definitions and analysis of velocity and acceleration, the statement is true. If an object has a constant velocity, that means its speed and direction are not changing with time, so the acceleration, which is the change in velocity, is indeed zero.

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!

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

(a) Show that the derivative of an odd function is even. That is, if \(f(-x)=-f(x),\) then \(f^{\prime}(-x)=f^{\prime}(x)\) (b) Show that the derivative of an even function is odd. That is, if \(f(-x)=f(x),\) then \(f^{\prime}(-x)=-f^{\prime}(x)\)

Use the position function \(s(t)=-16 t^{2}+v_{0} t+s_{0}\) for free-falling objects. A silver dollar is dropped from the top of a building that is 1362 feet tall. (a) Determine the position and velocity functions for the coin. (b) Determine the average velocity on the interval [1,2] . (c) Find the instantaneous velocities when \(t=1\) and \(t=2\). (d) Find the time required for the coin to reach ground level. (e) Find the velocity of the coin at impact.

In Exercises \(75-80\), evaluate the derivative of the function at the indicated point. Use a graphing utility to verify your result. \(\frac{\text { Function }}{f(x)=\frac{x+1}{2 x-3}} \quad \frac{\text { Point }}{(2,3)}\)

Existence of an Inverse Determine the values of \(k\) such that the function \(f(x)=k x+\sin x\) has an inverse function.

Determine the point(s) at which the graph of \(f(x)=\frac{x}{\sqrt{2 x-1}}\) has a horizontal tangent line.
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Calculus

Read Explanation

Mechanics Maths

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.