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
Last edited by Orac; 03/02/16 06:11 PM.
