motion

FROM s=ut +0.5 at^2

sn - sn-1 =

0.5a{n^2-(n-1)^2} -u =

0.5a(2n-1) - u=

a(n-0.5) - u

Correction

0.5a{n^2-(n-1)^2} + u =

0.5a(2n-1) + u=

a(n-0.5) + u

The starting velocity is u. The ending velocity after n seconds is u + na. This means the average velocity is:

[u + (u + na)] / 2 = u + na/2

Distance is average velocity x time so:

D = (u + na/2) x n = un + an^2/2

....."In the nth second" ?

......NB It's the area of the trapezium under the v-t graph in the nth second....

Thanks, I incorrectly read that as after n seconds.

You can use the same solution but move the starting conditions up to time n-1.

The velocity at n-1 seconds is u+ a(n-1) and the time is now one second, so you get

u + a(n-1) + a/2

which is what Fresco posted earlier.