Precalculus (6th Edition) Blitzer

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

Chapter 1 - Section 1.5 - More on Slope - Exercise Set - Page 226: 20


a. $84\ ft/sec$ b. $73\ ft/sec$ c. $72.12\ ft/sec$ d. $72.012\ ft/sec$

Work Step by Step

Given the distance equation $s(t)=12t^2$, we can evaluate the average velocity by calculating the ratio of displacement (change in distance) $\Delta s$ and the time interval $\Delta t$, that is $\bar v=\frac{\Delta s}{\Delta t}$ a. For $t_1=3, t_2=4$, we have $\Delta t=4-3=1$ and $\Delta s=s(4)-s(3)=12(4)^2-12(3)^2=84$, thus we have $\bar v=\frac{84}{1}=84\ ft/sec$ b. For $t_1=3, t_2=3.5$, we have $\Delta t=3.5-3=0.5$ and $\Delta s=s(3.5)-s(3)=12(3.5)^2-12(3)^2=39$, thus we have $\bar v=\frac{39}{0.5}=73\ ft/sec$ c. For $t_1=3, t_2=3.01$, we have $\Delta t=3.01-3=0.01$ and $\Delta s=s(3.01)-s(3)=12(3.01)^2-12(3)^2=0.7212$, thus we have $\bar v=\frac{0.7212}{0.01}=72.12\ ft/sec$ d. For $t_1=3, t_2=3.001$, we have $\Delta t=3.001-3=0.001$ and $\Delta s=s(3.001)-s(3)=12(3.001)^2-12(3)^2=0.072102$, thus we have $\bar v=\frac{0.072012}{0.001}=72.012\ ft/sec$
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