Explanation Given information: It is given that, s ( t ) = 1.9 t 2 . Concept used : Since average velocity of a function is in interval [t 0 , t 1 ] is: Avg velocity = Slope of secant line = s ( t 1 ) − s ( t 0 ) t 1 − t 0 Calculation: Since, s ( t ) = 22 t + 17 . Therefore average velocity of the function over the interval [ 5 , 9 ] is, Avg velocity = s ( 9 ) − s ( 5 ) 9 − 5 Avg velocity = 1.9 ( 9 ) 2 − 1.9 ( 5 ) 2 4 Avg velocity = 1.9 ( 81 − 25 ) 4 Avg velocity = 1.9 × 56 4 Avg velocity = 1 .9 × 14 Avg velocity = 26 .6 Now instantaneous velocity of the function at t = 5 is : Instt velocity of the function at [ t = 5 ] = Slope of tangent line at [ t = 5 ]

4th Edition

ISBN: 9781319279813

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 2.1, Problem 6E

To determine

To calculate:

The average velocity of the following function over the interval [5,9] :

s(t)=1.9t2.

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Chapter 2.1, Problem 5E

Chapter 2.1, Problem 7E

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Chapter 2 Solutions

CALCULUS:EARLY TRANS.-SAPLINGPLUS

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The average velocity of the following function over the interval [ 5 , 9 ] : s ( t ) = 1.9 t 2 .

Solution Summary: The author calculates the average velocity of the following function over the interval left[5,9right].

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To calculate:

The average velocity of the following function over the interval [5,9] :

s(t)=1.9t2.

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Chapter 2 Solutions

The average velocity of the following function over the interval [ 5 , 9 ] : s ( t ) = 1.9 t 2 .

Solution Summary: The author calculates the average velocity of the following function over the interval left[5,9right].

Concept explainers

Videos

To calculate: The average velocity of the following function over the interval [5,9] : s(t)=1.9t2.

Want to see the full answer?

Chapter 2 Solutions

To calculate:

The average velocity of the following function over the interval [5,9] :

s(t)=1.9t2.