Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 2 - Functions and Their Graphs - 2.1 Functions - 2.1 Assess Your Understanding - Page 59: 113

Answer

a) The stopping distance $D(v)=0.05v^2+2.6v-15$ b) $D(60)=321$ c) A vehicle travellng at a speed of $60 \text{ miles per hour}$ will need a stopping distance of $321$ feet.

Work Step by Step

Given, Reaction distance $R(v)=2.2v$ Breaking distance $B(v)=0.05v^2+0.4v-15$ a) The stopping distance is calculated as below $D(v)=R(v)+B(v)$ $=2.2v+0.05v^2+0.4v-15$ $=0.05v^2+2.6v-15$ b) Substitute $x=60$, To calculate the stopping distance at 60mph, $D(60)=0.05(60)^2+2.6(60)-15$ $=180+156-15$ $=321$ c) The car will need $321$ feet to stop once the impediment is observed.
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