Youe working in differentials so you want the average of the difference. The difference is 6 so yes the average differential is 3.

I think the issue is you went into differentials thinking it was going to help you ... it won't it makes it more abstract go back to you original form.

(i) v = u + at

v = your final velocity, u = your start velocity
a = your average acceleration, t = time

Plugging in some numbers so you see how it works. I start at 2 mph, my accel is 5mph/h and I accel for 3 hours

Your final velocity v = 2 + 5*3 = 17 mph .. I am sure you can see that is correct

(ii) Your average velocity Vavg = (v+u)/2

*Note the different formula for average because we aren't in differentials the bit that caused you grief

Ok substitute v from (i) into (ii)

(iii) Vavg = ((u+at)+u)/2

combine the u's

(iv) Vavg = (2u+at)/2

Simplify the divid by 2

(v) Vavg = u + 1/2at

As you correctly said velocity = distance /time and so you can rewrite Vavg with that formula

(vi) Vavg = Daccel / t

Substitute Vavg from (vi) back into (v)

(vii) Daccel/t = u + 1/2at

Remove the divid t from Daccel by taking to other side

(viii) Daccel = (u +1/2at) * t

simplify

(ix) Daccel = ut + 1/2at^2

So that is the distance the acceleration operated over lets check with our example above

Daccel = 2*3 + 1/2 * 5 * 3^2
= 6 + 1/2 * 5 * 9
= 6 + 22.5
= 28.5 miles

Check equation (v) and (vi) and you will see it is right

I think that is the distance relationship you were after


Edited by Orac (03/02/16 06:11 PM)
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I believe in "Evil, Bad, Ungodly fantasy science and maths", so I am undoubtedly wrong to you.