Explanation Given information: It is given that, s ( t ) = 2 t 3 . 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 ) = 2 t 3 . Therefore average velocity of the function over the interval [ 2 , 4 ] is, Avg velocity = s ( 4 ) − s ( 2 ) 4 − 2 Avg velocity = 2 ( 4 ) 3 − 2 ( 2 ) 3 2 Avg velocity = 2 ( 64 − 8 ) 2 Avg velocity = 2 × 56 2 Avg velocity = 56 Now instantaneous velocity of the function at t = 2 is : Instt velocity of the function at [ t = 2 ] = Slope of tangent line at [ t = 2 ]

4th Edition

ISBN: 9781319279813

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 2.1, Problem 9E

To determine

To calculate:

The average velocity of the following function over the interval [2,4] :

s(t)=2t3.

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Chapter 2.1, Problem 8E

Chapter 2.1, Problem 10E

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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 [ 2 , 4 ] : s ( t ) = 2 t 3 .

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To calculate: The average velocity of the following function over the interval [2,4] : s(t)=2t3.

Want to see the full answer?

Chapter 2 Solutions

To calculate:

The average velocity of the following function over the interval [2,4] :

s(t)=2t3.