You run two miles at an average speed of 8 mi/hr and then walk back at a rate of 4 mi/hr. What is the average speed for the trip? (to the nearest tenth)

Question

You run two miles at an speed of 8mi/hr and then walk back at a rate of 4 mi/hr.  What is the average speed for the trip?

Answer 1

distance=rate*time

2=8t running

2=4t walking

the time spent running is...

2=8t

2/8=t

1/4 hour=t running

the time spent walking is...

2=4t

2/4 hour=t walking

total distance=rate*total time

total distance is 2+2=4 miles

total time is 1/4 + 2/4=3/4 of an hour

total distance=rate*total time

4=rate*(3/4)

4=r(3/4)

multiply both sides by (4/3)

(4/3)(4)=r(3/4)(4/3)

(16/3)=r

r=5 1/3 miles per hour or 5.3 miles per hour is the average speed for the trip (not (8+4)/2=6 !!!)

another way to get the answer is to use what is called the harmonic mean if you understand it...

(2*8*4)/(8+4)=64/12=5 4/12=5 1/3 again

I wrote the harmonic mean in the simplest form so that you could understand it. If you are not familiar with it that's okay. It's a simple way of determining average speed over the same distance or average gas mileage over the same driving distance.

Answer 2

our average speed can be calculated in a couple different ways

 

is this case lets use a weighted average of the two speeds to find the average speed.

 

we know that that we were only going half as fast on the way back so it would take us twice as long

 

so our average speed would be (1/3) * 8 + (2/3) * 4

 

8/3 + 8/3 = 16/3  mph

 

another way to look at it is

 

to say we covered a total distance of 4 miles (2 out and 2 back)

 

the first 2 miles took us 1/4 of an hour since we were going at a rate of the 8 mph

 

the next 2 miles took us 1/2 of an hour since we were going at a rate of 4 mph

 

so we covered a distance of 4 miles in a time of 3/4 of an hour

 

distance = speed * time

 

speed = distance/time

 

speed = 4 /(3/4)

 

speed = 16/3  so either way we come up with the same answer