Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 4 - Integration - 4.1 Exercises - Page 253: 63


$a(t)=-\frac{1}{2t^\frac{3}{2}}$ $x(t) = 2\sqrt t+2$

Work Step by Step

Derivate $v(t)$ to find $a(t)$: $v(t)=\frac{1}{\sqrt t}$ $v'(t)=(-\frac{1}{2})t^{-\frac{1}{2}-1} = -\frac{1}{2t^\frac{3}{2}}$ $a(t)=v'(t)=-\frac{1}{2t^\frac{3}{2}}$ Integrate $v(t)$ to find $x(t)$: $v(t)=\frac{1}{\sqrt t}$ $\int f(t) dt=\int \frac{1}{\sqrt t} dt$ Apply power rule: $\int t^n dt = \frac{t^{n+1}}{n+1}$ with $n =-\frac{1}{2}$ $x(t)= 2\sqrt{t} + C$ At $t=1$, position have been given as $x=4$. Use this in the equation above to find C: $4= 2\sqrt{1} + C$ $C = 4 - 2\sqrt{1}=2$ Final equation: $x(t) = 2\sqrt t+2$
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