distance formula

popeyesmom

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I am having problems completing this math problem. Would you please explain in step by step instructions. Thank you.
Nolan Ryan was famous for pitching a baseball at speeds that exceeded 100 miles per hour. How long would a baseball take to travel 60 feet 6 inches--the distance from the pitcher's rubber to home plate--at a a speed of 100 miles per hour?
 
Hello, popeyesmom!

We need a procedure for converting units of measure.


Nolan Ryan was famous for pitching a baseball at speeds that exceeded 100 miles per hour.
How long would a baseball take to travel 60 feet 6 inches, the distance from the pitcher's rubber
to home plate, at a a speed of 100 miles per hour?

First, convert miles-per-hour to feet-per-second.

"100 miles per hour" means: .100 miles1 hour\displaystyle \dfrac{\text{100 miles}}{\text{1 hour}}

Since 1 mile=5280 feet\displaystyle \text{1 mile} = \text{5280 feet}, we can multiply by 5280 feet1 mile  \displaystyle \dfrac{\text{5280 feet}}{\text{1 mile}}\; which equals 1.

Since 1 hour=3600 seconds\displaystyle \text{1 hour} = \text{3600 seconds}, we can multiply by 1 hour3600 sec  \displaystyle \dfrac{\text{1 hour}}{\text{3600 sec}}\; which equals 1.


We have: .100  miles1  hour×5280 ft1  mile×1  hour3600  sec\displaystyle \dfrac{100\;\text{miles}}{1\;\text{hour}} \times \dfrac{5280\text{ ft}}{1\;\text{mile}} \times \dfrac{1\;\text{hour}}{3600\;\text{sec}}


Note how the units cancel:

. . 100  /////miles1  ////hour×5280 ft1  ////mile×1  ////hour3600  sec  =  1005280 ft3600 sec\displaystyle \dfrac{100\;\rlap{/////}{\text{miles}}}{1\;\rlap{////}\text{hour}} \times \dfrac{5280\text{ ft}}{1\;\rlap{////}\text{mile}} \times \dfrac{1\;\rlap{////}\text{hour}}{3600\;\text{sec}} \;=\;\dfrac{100\cdot5280\text{ ft}}{3600\text{ sec}}

Reduce the fraction: .44 ft3 sec=443 ft/sec\displaystyle \dfrac{44\text{ ft}}{3\text{ sec}} \:=\:\dfrac{44}{3}\text{ ft/sec}


Formula: .(Distance)=(Speed)×(Time)T=DS\displaystyle \text{(Distance)} \:=\:\text{(Speed)} \times \text{(Time)} \quad\Rightarrow\quad T \:=\:\dfrac{D}{S}

Therefore: .T  =  60.5 ft443 ft/sec  =  3380  =  0.4125 sec\displaystyle T \;=\;\dfrac{60.5\text{ ft}}{\frac{44}{3}\text{ ft/sec}} \;=\;\dfrac{33}{80} \;=\;0.4125\text{ sec}
 
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