Emerging briefly from the depths of my maths phobia, I permitted myself a couple of thoughts.

I found this. Acceleration is the time rate of change of velocity, not just change in velocity. This means that a does not equal dv. It means that a = dv/dt.

And thought: If v = l/t, then, in the equation a = dv/dt time occurs twice; on both occasions as t^-1. Is my thinking right?

Your thinking is correct but you mathematics is wrong


Close but not quite if you are using the differential form the it isn't "a" it's "da" or the differential acceleration so

da = dv/dt.

Remember whatever you do to one side of an equation in mathematics you must do to the other.

You took the differential of the right side then you must also take the differential of the left.

You also need to use care da, dv and dt are all averages if we assume uniform acceleration

dv is actually the average velocity between two differential points dvf (the final velocity) and dvi (the initial velocity). Fortunately time is constant.

Let me write the equations showing the averages

dv (average) = (dvf-dvi)/2

dt (average) = dt

da (average) = dv (average)/dt (average)

Thanks Orac, that seems to make sense, but some doubt creeps in when I introduce values.



If dvf = 10mph & dvi = 4mph, then (dvf-dvi)/2 = 3mph

Therefore, dv = 3mph?

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

Thanks again, a couple more runs through those equations and (hopefully) all will be clear.

I didn’t introduce differentials. Previously I had thought “a = dv” which seemed quite simple. Then I found:



This seemed to say my simplistic equation was wrong. While I was prepared to accept that; I found myself thinking: “v = l/t so this new equation has t^-1 twice on the RH side; what effect does that have?”

Sorry that was probably me then because you wrote dv, dt etc I assumed you wanted to use differentials.

My bad.

You know how some scientists are afraid of absolute infinity, in case it lets God into the discussion. Well, to me the word "differential" carries the threat of letting calculus in.

I am the reverse where there be calculus there be an excuse for beer ... or rum if ye is a pirate.

Perhaps a few beers, or some single malt, would help me with calculus. I think it might take a lot, though.