Question

Travis jogged to his friend's house 10 miles away and then got a ride back home. It took him 2 hours longer to jog there than ride back. His jogging rate was 16 mph slower than the rate when he was riding. What was his jogging rate?

Answer asap please :)))

Answer 1

Using the relation between velocity, distance and time, it is found that his jogging rate was of 4.5 mph.

What is the relation between velocity, distance and time?

Velocity is distance divided by time, hence:

$v = \frac{d}{t}$

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

$v = \frac{10}{t + 2}$

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

$v + 16 = \frac{10}{t}$

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

$\frac{10}{t + 2} + 16 = \frac{10}{t}$

$\frac{10}{t} - \frac{10}{t + 2} = 16$

$\frac{10t + 20 - 10t}{t(t + 2)} = 16$

16t² + 32t - 20 = 0 -> t = 0.5.

Then his jogging rate in mph was:

$v = \frac{10}{0.5 + 2} = 4.5$

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

#SPJ1