Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 11 - Review Exercises - Page 1180: 60


a. $ 16\ ft/sec$, $ -48\ ft/sec$ b. $2.5\ sec$, $105\ ft$

Work Step by Step

a. We can calculate the instantaneous velocity as the instantaneous rate of change from the height formula as $v(t)=s'(t)=lim_{h\to0}\frac{s(t+h)-s(t)}{h}=lim_{h\to0}\frac{-16(t+h)^2+80(t+h)+5-(-16t^2+80t+5)}{h}=lim_{h\to0}\frac{-32ht-16h^2+80h}{h}=lim_{h\to0}(-32t-16h+80)=-32t+80$. Thus we have $v(2)=-32(2)+80=16\ ft/sec$ and $v(4)=-32(4)+80=-48\ ft/sec$ b. Let $v(t)=0$. We have $t=\frac{80}{32}=2.5\ sec$ and $s(2.5)=-16(2.5)^2+80(2.5)+5=105\ ft$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.