Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 2 - Limits - 2.1 The Idea of Limits - 2.1 Exercises - Page 59: 10

Answer

$\mathrm{a}.\qquad 15.3$ $\mathrm{b}.\qquad 20.2$ $\mathrm{c}.\qquad 25.1$ $\mathrm{d}.\qquad -4.9h+30$

Work Step by Step

$\mathrm{a}$. Over $[0,3]$, $v_{\mathrm{a}\mathrm{v}}=\displaystyle \frac{s(3)-s(0)}{3-0}=\frac{65.9-20}{3}=15.3$. $\mathrm{b}$. Over $[0,2]$, $v_{\mathrm{a}\mathrm{v}}=\displaystyle \frac{s(2)-s(0)}{2-0}=\frac{60.4-20}{2}=20.2$. $\mathrm{c}$. Over $[0,1]$, $v_{\mathrm{a}\mathrm{v}}=\displaystyle \frac{s(1)-s(0)}{1-0}=\frac{45.1-20}{1}=25.1$. $\mathrm{d}$. Over $[0, h]$, $v_{\mathrm{a}\mathrm{v}}=\displaystyle \frac{s(h)-s(0)}{h-0}$ $=\displaystyle \frac{(-4.9h^{2}+30h+20)-20}{h}$ $=\displaystyle \frac{h(-4.9h+30)}{h}\qquad$ ... h cancels ... $=-4.9h+30$.
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