Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.1 - Tangents and the Derivative at a Point - Exercises 3.1: 29

Answer

$v(2)=-19.6$

Work Step by Step

This is a kinematics question. A function is given to us to represent the height as a function of time. $h(t)=100-4.9t^2$ Given that the question is asking how fast is the object falling at t=2, it might be advantageous for us to find the instantaneous rate of change of that function at that time, ie. the velocity. $h'(t)=-(2)(4.9)t=-9.8t$ $h'(2)=-9.8(2)=-19.6$ The object is thus falling down with a velocity of 19.6 meters per second at a time of two seconds.
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