NO, it is not the average of the two speeds, because the times are different. Look:

Assume the distance for the whole trip is 96 miles.

Let’s say the first half is covered at a rate of 24 mph. That’s 48 miles/ 24 mph to get a time of 2 hours.

Let’s say the second half is covered at a rate of 16 mph. That’s 48/16 = 3 hours.

NOW, the whole distance is 96 miles, and the whole time is 5 hours. So the average speed for the whole trip is 96/5, about 19.2 mph. That’s NOT the average of 16 and 24.

I’ve taught math for years at the college level. Averaging the rates is a typical error for this kind of problem. Don’t make that mistake!

Finding the average is a very simple exercise.

Simply add the two speed then divide it by two.

In other instances, however many items you add you must divide by that number. For e.g. to find the average of these numbers: 12, 18, 30, I do this:

(12+18+30)/3=60/3

=20

let me give you the general formula:

v(average) = (m1*v1 + m2*v2) / (m1 + m2)

where m1 is the distance covered with velocity v1 and m2 with velocity v2. I belive this can be extended to any number of velocities.

Here m1 = m2 since first half = second half.

Thus formula becomes (v1 + v2)/2.

So that’s your answer (v1 + v2)/2.

(v1+v2)/2= average speed

Uhm, gee…this is a stressful one! permit’s see: 20 + 60 = eighty, divide via 2 to get primary of…look ahead to it…40mph. EDIT: Brian, i could be incorrect, i will carry out a little checking, yet i did no longer supply you the thumbs down. besides, i’ve got accomplished greater suitable: i’ve got have been given 2 thumbs down. So there…! 2d EDIT: I concede. I remodeled the difficulty utilising an historical mathmatical equation from my college days and wide-unfold that Brian’s answer of 30mph is right. I won’t define the equation, because of the fact its simplicity while in comparison with that shown via Brian could make modern-day-day maths instructors squirm with embarrassment. Te he!

yeah, like the guy above me said

(V1+V2)/2

assume total distance=x

now, time taken for the first half=dist/speed=x/2v1

similarly, time for 2nd half=x/2v2

so, total itme=x/2v1+x/2v2= (xv2+xv1)/2v1v2=x(v1+v2)/2v1v2

so, av. speed =dist./time=x/x(v1+v2)/2v1v2=2v1v2/(v1+v2)

ANSWER;-2V1V2/(V1+V2)

as above