Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 2 - Section 2.1 - Functions - 2.1 Exercises - Page 158: 81

Answer

(a) Respectively: 8.66 m, 6.61 m and 4.36 (b) The length of an object decreases as its velocity increases.

Work Step by Step

(a) Substitute the values for $v$ into the function and use a calculator to find $L(v)$ in each case. $$L(0.5c) = 10 \sqrt {1 - \frac {(0.5c)^2}{c^2}} = 10 \sqrt {1 - \frac {0.5^2c^2}{c^2}}$$ $$L(0.5c) = 10 \sqrt {1 - 0.5^2} = 8.66 $$ $$L(0.75c) = 10 \sqrt {1 - \frac {(0.75c)^2}{c^2}} = 10 \sqrt {1 - 0.75^2} $$ $$L(0.75c) = 6.61$$ $$L(0.9c) = 10 \sqrt {1 - \frac {(0.9c)^2}{c^2}} = 10 \sqrt {1 - 0.9^2} $$ $$L(0.9c) = 4.36$$ (b) As we can see, the length of the object decreased from 8.66 to 4.36 when $v$ was increased from 0.5 to 0.9 Thus, the length of an object decreases as its velocity increases.
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