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High School Math Calculus Advanced Placement Calculus Graphical Numerical Algebraic Sixth Edition High School Binding Copyright 2020

The value of time when the particle moves forward, backward, increase in speed, decrease in speed.

The value of time when the particle moves forward, backward, increase in speed, decrease in speed.

Solution Summary: The author calculates the value of time when the particle moves forward, backward, increase in speed, and decrease in velocity.

Advanced Placement Calculus Graphical Numerical Algebraic Sixth Edition High School Binding Copyright 2020

Advanced Placement Calculus Graphical Numerical Algebraic Sixth Edition High School Binding Copyright 2020

6th Edition

ISBN: 9781418300203

Author: Prentice Hall

Publisher: Prentice Hall

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Question

Chapter 2.4, Problem 9E

a.

To determine

To calculate: The value of time when the particle moves forward, backward, increase in speed, decrease in speed.

a.

Expert Solution

Answer to Problem 9E

The particle moves forward from t=0 sec to t=1 sec , from t=5 sec to t=6 sec, moves backward from t=1 sec to t=5 sec , speed of particle increase from t=3 sec to t=6 sec and speed of particle decreases from t=0 sec to t=2 sec, t=6 sec to t=7 sec.

Explanation of Solution

Given information:

The graph of the velocity verses time

Advanced Placement Calculus Graphical Numerical Algebraic Sixth Edition High School Binding Copyright 2020, Chapter 2.4, Problem 9E , additional homework tip 1

Concept used:

The displacement of the particle is final distance minus initial distance.

Calculation:

When the particle is in positive y axis then the particle will move forward and when the particle is in negative y axis the particle will move backwards.

In the given graph, the particle is in positive y axis from t=0 sec to t=1 sec , from t=5 sec to t=6 sec.

In the given graph, the particle is in positive y axis from t=1 sec to t=5 sec.

When the slope is positive that is the line is moving upward from left to right then the speed of the particle will increase and if the slope is negative that is line is moving downwards from left to right then the speed of the particle will decrease.

The slope of the line is positive from t=3 sec to t=6 sec, the speed of the particle increases.

The slope of the line is negative from t=0 sec to t=2 sec, t=6 sec to t=7 sec, the speed of the particle decreases.

Conclusion: The particle moves forward from t=0 sec to t=1 sec , from t=5 sec to t=6 sec, moves backward from t=1 sec to t=5 sec , speed of particle increase from t=3 sec to t=6 sec and speed of particle decreases from t=0 sec to t=2 sec, t=6 sec to t=7 sec.

b.

To determine

To calculate: The time when acceleration is positive, negative and constant.

b.

Expert Solution

Answer to Problem 9E

The time of positive acceleration is from t=3 sec to t=6 sec, negative acceleration from t=0 sec to t=2 sec, t=6 sec to t=7 sec and zero acceleration from t=2 sec to t=3 sec and from t=7 sec to t=9 sec.

Explanation of Solution

Given information:

The graph of the velocity verses time

Advanced Placement Calculus Graphical Numerical Algebraic Sixth Edition High School Binding Copyright 2020, Chapter 2.4, Problem 9E , additional homework tip 2

Calculation:

When the slope is positive, then the acceleration will be positive.

The acceleration of the particle is positive from t=3 sec to t=6 sec.

When the slope is negative, then the acceleration will be negative.

The acceleration of the particle is negative from t=0 sec to t=2 sec, t=6 sec to t=7 sec.

The acceleration of the particle will be zero when the velocity is constant.

So the acceleration will be constant from t=2 sec to t=3 sec and from t=7 sec to t=9 sec.

Conclusion: The time of positive acceleration is from t=3 sec to t=6 sec, negative acceleration from t=0 sec to t=2 sec, t=6 sec to t=7 sec and zero acceleration from t=2 sec to t=3 sec and from t=7 sec to t=9 sec.

c.

To determine

To calculate: The greatest speed of the particle.

c.

Expert Solution

Answer to Problem 9E

The value of time is t=0 sec and from t=2 sec to t=3 sec.

Explanation of Solution

Given information:

The graph of velocity verses time is shown below.

Advanced Placement Calculus Graphical Numerical Algebraic Sixth Edition High School Binding Copyright 2020, Chapter 2.4, Problem 9E , additional homework tip 3

Concept used:

The speed is always a positive quantity.

Calculation:

The speed is greatest at t=0 sec and from t=2 sec to t=3 sec.

Conclusion: The value of time is t=0 sec and from t=2 sec to t=3 sec.

d.

To determine

To calculate:. The value of time

d.

Expert Solution

Answer to Problem 9E

The time is from t=7 sec to t=9 sec.

Explanation of Solution

Given information:

The function is st=t2−3t+2 and time is t=4

Calculation:

The particle is standing still from t=7 sec to t=9 sec.

Conclusion: The time is from t=7 sec to t=9 sec.

Chapter 2.4, Problem 8E

Chapter 2.4, Problem 10E

Chapter 2 Solutions

Advanced Placement Calculus Graphical Numerical Algebraic Sixth Edition High School Binding Copyright 2020

Ch. 2.1 - Prob. 1QR Ch. 2.1 - Prob. 2QR Ch. 2.1 - Prob. 3QR Ch. 2.1 - Prob. 4QR Ch. 2.1 - Prob. 5QR Ch. 2.1 - Prob. 6QR Ch. 2.1 - Prob. 7QR Ch. 2.1 - Prob. 8QR Ch. 2.1 - Prob. 9QR Ch. 2.1 - Prob. 10QR

Ch. 2.1 - Prob. 1E Ch. 2.1 - Prob. 2E Ch. 2.1 - Prob. 3E Ch. 2.1 - Prob. 4E Ch. 2.1 - Prob. 5E Ch. 2.1 - Prob. 6E Ch. 2.1 - Prob. 7E Ch. 2.1 - Prob. 8E Ch. 2.1 - Prob. 9E Ch. 2.1 - Prob. 10E Ch. 2.1 - Prob. 11E Ch. 2.1 - Prob. 12E Ch. 2.1 - Prob. 13E Ch. 2.1 - Prob. 14E Ch. 2.1 - Prob. 15E Ch. 2.1 - Prob. 16E Ch. 2.1 - Prob. 17E Ch. 2.1 - Prob. 18E Ch. 2.1 - Prob. 19E Ch. 2.1 - Prob. 20E Ch. 2.1 - Prob. 21E Ch. 2.1 - Prob. 22E Ch. 2.1 - Prob. 23E Ch. 2.1 - Prob. 24E Ch. 2.1 - Prob. 25E Ch. 2.1 - Prob. 26E Ch. 2.1 - Prob. 27E Ch. 2.1 - Prob. 28E Ch. 2.1 - Prob. 29E Ch. 2.1 - Prob. 30E Ch. 2.1 - Prob. 31E Ch. 2.1 - Prob. 32E Ch. 2.1 - Prob. 33E Ch. 2.1 - Prob. 34E Ch. 2.1 - Prob. 35E Ch. 2.1 - Prob. 36E Ch. 2.1 - Prob. 37E Ch. 2.1 - Prob. 38E Ch. 2.1 - Prob. 39E Ch. 2.1 - Prob. 40E Ch. 2.1 - Prob. 41E Ch. 2.1 - Prob. 42E Ch. 2.1 - Prob. 43E Ch. 2.1 - Prob. 44E Ch. 2.1 - Prob. 45E Ch. 2.1 - Prob. 46E Ch. 2.2 - Prob. 1QR Ch. 2.2 - Prob. 2QR Ch. 2.2 - Prob. 3QR Ch. 2.2 - Prob. 4QR Ch. 2.2 - Prob. 5QR Ch. 2.2 - Prob. 6QR Ch. 2.2 - Prob. 7QR Ch. 2.2 - Prob. 8QR Ch. 2.2 - Prob. 9QR Ch. 2.2 - Prob. 10QR Ch. 2.2 - Prob. 1E Ch. 2.2 - Prob. 2E Ch. 2.2 - Prob. 3E Ch. 2.2 - Prob. 4E Ch. 2.2 - Prob. 5E Ch. 2.2 - Prob. 6E Ch. 2.2 - Prob. 7E Ch. 2.2 - Prob. 8E Ch. 2.2 - Prob. 9E Ch. 2.2 - Prob. 10E Ch. 2.2 - Prob. 11E Ch. 2.2 - Prob. 12E Ch. 2.2 - Prob. 13E Ch. 2.2 - Prob. 14E Ch. 2.2 - Prob. 15E Ch. 2.2 - Prob. 16E Ch. 2.2 - Prob. 17E Ch. 2.2 - Prob. 18E Ch. 2.2 - Prob. 19E Ch. 2.2 - Prob. 20E Ch. 2.2 - Prob. 21E Ch. 2.2 - Prob. 22E Ch. 2.2 - Prob. 23E Ch. 2.2 - Prob. 24E Ch. 2.2 - Prob. 25E Ch. 2.2 - Prob. 26E Ch. 2.2 - Prob. 27E Ch. 2.2 - Prob. 28E Ch. 2.2 - Prob. 29E Ch. 2.2 - Prob. 30E Ch. 2.2 - Prob. 31E Ch. 2.2 - Prob. 32E Ch. 2.2 - Prob. 33E Ch. 2.2 - Prob. 34E Ch. 2.2 - Prob. 35E Ch. 2.2 - Prob. 36E Ch. 2.2 - Prob. 37E Ch. 2.2 - Prob. 38E Ch. 2.2 - Prob. 39E Ch. 2.2 - Prob. 40E Ch. 2.2 - Prob. 41E Ch. 2.2 - Prob. 42E Ch. 2.2 - Prob. 43E Ch. 2.2 - Prob. 44E Ch. 2.2 - Prob. 45E Ch. 2.2 - 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Prob. 46RE Ch. 2 - Prob. 47RE Ch. 2 - Prob. 48RE Ch. 2 - Prob. 49RE Ch. 2 - Prob. 50RE Ch. 2 - Prob. 51RE Ch. 2 - Prob. 52RE Ch. 2 - Prob. 53RE Ch. 2 - Prob. 54RE Ch. 2 - Prob. 55RE Ch. 2 - Prob. 56RE Ch. 2 - Prob. 57RE Ch. 2 - Prob. 58RE Ch. 2 - Prob. 59RE Ch. 2 - Prob. 60RE Ch. 2 - Prob. 61RE Ch. 2 - Prob. 62RE Ch. 2 - Prob. 63RE Ch. 2 - Prob. 64RE Ch. 2 - Prob. 65RE Ch. 2 - Prob. 66RE Ch. 2 - Prob. 67RE Ch. 2 - Prob. 68RE Ch. 2 - Prob. 69RE Ch. 2 - Prob. 70RE Ch. 2 - Prob. 71RE Ch. 2 - Prob. 72RE Ch. 2 - Prob. 73RE Ch. 2 - Prob. 74RE Ch. 2 - Prob. 75RE Ch. 2 - Prob. 76RE Ch. 2 - Prob. 77RE Ch. 2 - Prob. 78RE Ch. 2 - Prob. 79RE Ch. 2 - Prob. 80RE Ch. 2 - Prob. 81EP Ch. 2 - Prob. 82EP Ch. 2 - Prob. 83EP

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