Answer:
The average speed was 10.12 m/s or 22.96 mi/h
Explanation:
The average speed is defined as:
[tex]v = \frac{d}{t}[/tex] (1)
Where d is the total distance traveled and t is the passed time.
For the special case of Joseph DeLoach, he traveled a total distance of 200 meters in 19.75 seconds. Those values can be introduced in equation 1:
[tex]v = \frac{200 m}{19.75 s}[/tex]
[tex]v = 10.12 m/s[/tex]
That means that Joseph DeLoach traveled a distance of 10.12 meters per second.
To represent the result in miles per hour, it is necessary to know that 1 mile is equivalent to 1609 meters.
[tex]200 m x \frac{1 mi}{1609 m}[/tex] ⇒ 0.124 mi
it is needed to express the given time in units of hour. 1 hour is equivalent to 3600 seconds.
[tex]19.75 s x \frac{1 h}{3600 s}[/tex] ⇒ 0.0054 h
Then, equation 1 is used with the new representation of the values.
[tex]v = \frac{0.124 mi}{0.0054 h}[/tex]
[tex]v = 22.96 mi/h[/tex]
