Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 1 - Section 1.10 - Modeling with Functions - Exercise Set - Page 291: 3

Answer

See the full explanation below.

Work Step by Step

(a) Consider the provided statement “In $1954$, Roger Bannister of Britain cracked the $4-\text{minute}$ mark, setting the record for running a mile in 3 minutes, 59.4 seconds, or 239.4 seconds. In the half century then, the record has decreased by $0.3$ seconds per year.” Thus, to express the record time for the mile run, the linear function representing the record time for a mile run will be of the form $M\left( x \right)=mx+b$. Suppose that running x miles will consume equal time in total for a mile run after $1954$. Therefore, the record time for setting the record for running a mile in 239.4 seconds after running x miles and a year which is decreased by $0.3$ second per year is: $239.4-\left( 0.3 \right)x$. Thus, by above given conditions $M\left( x \right)=239.4-0.3x$ Hence, to express the record time for the mile run: $M\left( x \right)=239.4-0.3x$. (b) Consider the provided statement “In $1954$, Roger Bannister of Britain cracked the $4-\text{minute}$ mark, setting the record for running a mile in 3 minutes, 59.4 seconds, or 239.4 seconds. In the half century then, the record has decreased by $0.3$ seconds per year.” Thus, to calculate the year in which someone will run a $3-\text{minute}$, or $180-\text{second}$ mile use the linear function representing the record time for a mile run of the form $M\left( x \right)=239.4-0.3x$ from part (a). Therefore, from the above given condition. $\begin{align} & M\left( x \right)=239.4-0.3x \\ & 180=239.4-0.3x \\ & 59.4=0.3x \\ & 198=x \end{align}$ Hence, the year in which someone will run a $3-\text{minute}$ , or $180-\text{second}$ mile is $198\text{ years}$ after 1954, in 2152.
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