A.M 1999 Q.1 b(iii)

Shox

have it all up to this! keep hitting dead ends!

cocoa

hmm, weird question but anyway...

seriously shox, could you please give more material when you're asking a question? I mean, not only did I have to go and find my copy of the exam papers, but now I have to work out part b ii despite the fact that you have no problems with it apparently... In future it would be cool if you could outline the question yourself instead of pointing towards it...

anyway...

part ii :
u + 2ft = v
v=55/5t=11/t
u+2ft+11/t u = 11/t - 2ft
15 = 2ut + 2ft^2
2ft = 15/t - 2u
u=11/t -15/t + 2u
u=4/t

part iii:
clearly the only problem is finding the distance traveled in the last portion of the journey. I'm just going to write an equation for that final distance (the one we want) and then swap in values for everything in t (you could, hypothetically, use values of something else, f maybe, but as a few things are already valued in t, it seems easier this way)
(I'll use T for this final time, to avoid confusion)
s = ut + at^2/2 = uT - 3/2 fT^2
we know quite easily that u=11/t
s=11T/t -3/2 fT^2
to find T:
v = 0 =u + at = 11/t - 3fT
T=11/3ft
s= 11/t * 11/3ft - 3/2 * f * 121/9f^2t^2
=121/3ft^2 - 121/6ft^2
=121/6ft^2
to find f:
15 = 2ut + 2ft^2 (u in this case is 4/t)
f = 15/2t^2 - u/t = 15/2t^2 - 4/t^2 = 7/2t^2
f=7/2t^2
s = 121/6t^2 * 2t^2/7 = 121/21 = 5.76m
now we simply add the first two distances (15m and 55m) to that and we have 75.76m

Shox

ok sry cocoa, your absolutey right. I'll post more info from now on. tanx again for the great help!