College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Section 1.4 - Quadratic Equations - 1.4 Exercises - Page 113: 75



Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the properties of equality to solve the given equation, $ r=r_0+\dfrac{1}{2}at^2 ,$ for $ t .$ $\bf{\text{Solution Details:}}$ Using the properties of equality, the equation above is equivalent to \begin{array}{l}\require{cancel} r-r_0=\dfrac{1}{2}at^2 \\\\ 2(r-r_0)=\left( \dfrac{1}{2}at^2 \right)2 \\\\ 2(r-r_0)=at^2 \\\\ \dfrac{2(r-r_0)}{a}=t^2 \\\\ t^2=\dfrac{2(r-r_0)}{a} .\end{array} Taking the square root of both sides (Square Root Principle) results to \begin{array}{l}\require{cancel} t=\pm\sqrt{\dfrac{2(r-r_0)}{a}} .\end{array} Multiplying the radicand by an expression equal to $1$ which will make the denominator a perfect power of the index results to \begin{array}{l}\require{cancel} t=\pm\sqrt{\dfrac{2(r-r_0)}{a}\cdot\dfrac{a}{a}} \\\\ t=\pm\sqrt{\dfrac{2a(r-r_0)}{a^2}} \\\\ t=\pm\sqrt{\dfrac{1}{a^2}\cdot2a(r-r_0)} \\\\ t=\pm\sqrt{\left( \dfrac{1}{a} \right)^2\cdot2a(r-r_0)} \\\\ t=\pm\left| \dfrac{1}{a} \right|\sqrt{2a(r-r_0)} .\end{array} Assuming that all variables are positive, the equation above is equivalent to \begin{array}{l}\require{cancel} t=\pm\dfrac{1}{a}\sqrt{2a(r-r_0)} \\\\ t=\pm\dfrac{\sqrt{2a(r-r_0)}}{a} .\end{array}
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.