1. uniform acceleration

A particle travels srating with a initial speed u, with uniform acceleration a. Show that the distance travelled during the nth second is u+an-.5a. (b) if the particle travels 17m in the 2nd second of motion and 47m in the 7th second of motion how far will it go in the (i) 10th second of motion (ii) nth second of motion

i worked out the first bit by letting s=un+.5an^2. i then found s=u(n+1)=5a(n+1)^2. and i took these away from each other and i was left with the answer. i no not know what to do for part b though

2. Re: uniform acceleration

what you've done isn't quite right

the first second is from 0 to 1, the 2nd from 1 to 2, etc.

so the $nth$ second is $t\in [n-1,n)$

putting this into the equation of motion we get

$d(n) = \dfrac{a}{2}(n^2) + u n - \dfrac{a}{2}(n-1)^2 - u(n-1) = \dfrac{a}{2}(n^2) + u n - \dfrac{a}{2}(n^2 - 2n + 1)- u n + u$

$d(n) = a\left(n-\dfrac 1 2\right)+u$

Now for (b) we note

$d(2) = 17$

$d(7) = 47$

$\dfrac{3a}{2} + u = 17$

$\dfrac{13a}{2} + u = 47$

Solving this we get

$5a=30$

$a=6,~u=8$

$d(10)=6(9.5)+8 = 65$

$d(n) = 6\left(n-\dfrac 1 2 \right) + 8$

3. Re: uniform acceleration

If you did what you said, you should have ended up with u + an + 0.5a rather than u + an - 0.5a.

The first second of motion happens between n=0 and n=1.
The second second of motion happens between n=1 and n=2, etc.

So the nth second of motion happens between n-1 and n (not n and n+1 which is what you have done)

So, first of all, you need to fix up part (a).

Part (b) uses the fact that the distance travelled during the nth second is u + an - 0.5a.

So, when n = 2, u+an-0.5a = u+2a-0.5a = u+1.5a = 17

and when n=7 , u+an-0.5a = ..... etc

This gives 2 simultaneous equations so you can work out u and a. Then sub into the distance formula u+an-0.5a to get a formula in terms of n only. Then you can do (i) and (ii).