Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 1 - Section 1.5 - Problem Solving and Using Formulas - Exercise Set: 67

Answer

$v=\dfrac{-at^2+2s}{2t}$

Work Step by Step

Subtract $\frac{1}{2}at^2$ from both sides to find: $s-\frac{1}{2}at^2=\frac{1}{2}at^2 +vt - \frac{1}{2}at^2 \\s-\frac{1}{2}at^2=vt$ Divide both sides by $t$ to find: $\dfrac{s-\frac{1}{2}at^2}{t}=\dfrac{vt}{t} \\\dfrac{s-\frac{1}{2}at^2}{t}=v \\\dfrac{\frac{2s}{2} - \frac{at^2}{2}}{t}=v \\\dfrac{\frac{2s-at^2}{2}}{t}=v \\\dfrac{2s-at^2}{2t}=v \\v=\dfrac{-at^2+2s}{2t}$
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