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:

Jogging, his velocity was of v, while the time was of t + 2, for a distance of 10 miles, hence:

Riding, his velocity was of 16, while the time was of t, also for a distance of 10 miles, hence:

Then, considering the first equation and replacing in the second:



16t² + 32t - 20 = 0 -> t = 0.5.
Then his jogging rate in mph was:

More can be learned about the relation between velocity, distance and time at brainly.com/question/24316569
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