Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.1 Derivatives and Rates of Change - 2.1 Exercises: 11

Answer

a.) From $t=0$ to $t=1$ the particle moves $3$ meters to the right. $\\$ From $t=1$ to $t=2$ the particle stands still. $\\$ From $t=2$ to $t=3$ the particle moves $2$ meters to the left. $\\$ From $t=3$ to $t=4$ the particle stands still. $\\$ From $t=4$ to $t=5$ the particle moves $1$ meter to the right. $\\$ From $t=5$ to $t=6$ the particle moves $1$ meter to the right. $\\$ b.)
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Work Step by Step

a.) The slope of the graph represents the velocity of the particle. If the slope is positive, it means that the particle is moving to the right. If the slope is negative, it means that the particle is moving to the left. And if the slope is 0 (the graph represents an horizontal line), it means that the particle is standing still. b.) From $t=1$ to $t=2,$ the particle moves to the right (thus it has a positive velocity) with a velocity of $3 m/s$ From $t=1$ to $t=2,$ the particle is standing still, therefore it has a velocity of $0 m/s$ From $t=2$ to $t=3$ the particle moves to the left (thus it has a negative velocity) with a velocity of $2 m/s$ From $t=3$ to $t=4,$ the particle is standing still, therefore it has a velocity of $0 m/s$ From $t=4$ to $t=6,$ the particle moves to the right (thus it has a positive velocity) with a velocity of $1 m/s$
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