Thomas' Calculus 13th Edition

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

Chapter 3: Derivatives - Section 3.4 - The Derivative as a Rate of Change - Exercises 3.4 - Page 133: 9

Answer

$7.5$ sec, $1.2$ sec.

Work Step by Step

Step 1. Identify the equations: $s(t)=1.86t^2$ on Mars, $s(t)=11.44t^2$ on Jupiter. Step 2. On Mars, $v(t)=s'(t)=2\times1.86t=3.72t$. Let $v(t)=27.8$; we have $3.72t=27.8$, which gives $t\approx7.5$ sec. Step 3. On Jupiter, $v(t)=s'(t)=2\times11.44t=22.88t$. Let $v(t)=27.8$; we have $22.88t=27.8$, which gives $t\approx1.2$ sec.
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