Math homework help algebra

i know the answer but i need help how 2 get there.
Emmett runs once around the school track at an average rate of 8 miles per hour. Then he jogs once around the track at an average rate of 6 miles per hour. what is his average rate for the two laps of the track? Use formula Distance=rate*time (answer 7.5mph)

These really trip people up for sure.....easy, but conceptually tricky.

You have d = rt

==> r1 = d1/t1 and d2 = r2/t2


But you need the total distance and the total time. The distances are equal, so d1 = d2 = d

So the trip is twice around the track....2d

=> avg speed = 2d/(t1 + t2), but t1 = d/r1 and t2 = d/r2
=> avg speed = 2d/(d/r1 + d/r2)
=> avg speed = 2/(1/r1 + 1/r2)

This simplifies to: 2r1r2/(r1 + r2)

r1 = 8
r2 = 6

Then your average is: 2*8*6/(8 + 6) = 6.86 [units]

I don't know how you're getting 7.5....I'm pretty confident about what I have shown you. This is often referred to as a harmonic mean and it's classic to use it in discussing average speeds, but assure you it is not just simply adding the two speeds and dividing by 2...no no no....lol

First, we need to assume how long the track is. It doesn't matter what length you assume, you will get the same answer. Let's assume it's a 1/4 mile track or .25 miles.

Use that to figure out how long it takes him to travel the track twice.

First lap
Distance = rate x time
0.25= 8t
t=0.03125 hours

Second lap
Distance = rate x time
0.25=6t
t=0.04167

Total time = 0.03125 hours +0.04167 =0.07292 hours
Total distance = .25 + .25 = 0.5 miles

Average rate = 0.5 miles/0.07292 hours = 6.857 = 6.9 mph

(8 + 6)/2 = 7mph
If the number of laps at each speed were different, you'd use a weighted average. For example, if he ran 3 laps at 8mph and 1 lap at 6mph, then his average would be