Using the relation between velocity, distance and time, it is found that his jogging rate was of 4.5 mph.
<h3>What is the relation between velocity, distance and time?</h3>
Velocity is <u>distance divided by time,</u> hence:
![v = \frac{d}{t}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7Bd%7D%7Bt%7D)
Jogging, his velocity was of v, while the time was of t + 2, for a distance of 10 miles, hence:
![v = \frac{10}{t + 2}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7B10%7D%7Bt%20%2B%202%7D)
Riding, his velocity was of 16, while the time was of t, also for a distance of 10 miles, hence:
![v + 16 = \frac{10}{t}](https://tex.z-dn.net/?f=v%20%2B%2016%20%3D%20%5Cfrac%7B10%7D%7Bt%7D)
Then, considering the first equation and replacing in the second:
![\frac{10}{t + 2} + 16 = \frac{10}{t}](https://tex.z-dn.net/?f=%5Cfrac%7B10%7D%7Bt%20%2B%202%7D%20%2B%2016%20%3D%20%5Cfrac%7B10%7D%7Bt%7D)
![\frac{10}{t} - \frac{10}{t + 2} = 16](https://tex.z-dn.net/?f=%5Cfrac%7B10%7D%7Bt%7D%20-%20%5Cfrac%7B10%7D%7Bt%20%2B%202%7D%20%3D%2016)
![\frac{10t + 20 - 10t}{t(t + 2)} = 16](https://tex.z-dn.net/?f=%5Cfrac%7B10t%20%2B%2020%20-%2010t%7D%7Bt%28t%20%2B%202%29%7D%20%3D%2016)
16t² + 32t - 20 = 0 -> t = 0.5.
Then his jogging rate in mph was:
![v = \frac{10}{0.5 + 2} = 4.5](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7B10%7D%7B0.5%20%2B%202%7D%20%3D%204.5)
More can be learned about the relation between velocity, distance and time at brainly.com/question/24316569
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