Mechanics Help!!! PLZ!!!

[IsOLaTiOnIsT]

kk i got some q's on mechanics, and im useless at mechanics, or some parts of mechanics. i really need the help as i got about 3 q's to do, i'd like to say thanks in advance if u can solve them, as i've already and still am trying:mad: .

1 Two particles, X and Y are moving in the same direction on parallel horizontal tracks. At a certain point O, the particle X, travelling with a speed of 16m/s and retarding uniformaly at 6m/s, overtakes Y, which is travelling at 8m/s and accelerating uniformaly at 2m/s. Calculate:

i - the distance of Y from O when the velocities of X and Y are equal,
ii - the velocity of X when Y overtakes X

2 a - A particle is projected vertically upwards with velocity 12m/s. Calculate:

i - the greatest height reached,
ii - its speed when it is at a height of 4m

b/Two Particles A and B are 96m apart on a smooth horizontal surface. A is moving directly towards B with speed 7m/s and retardation 2m/s. B is moving directly towards A with speed with 3m/s and accelerating 4m/s. Calculate the time taken before they meet

3/When t=0 (where t denotes time in seconds) a particle A moves from a point O along a straight line with inital velocity u m/s and constant acceleration a m/s. When t=4, a particle B moves from O along the straight line with inital velocity ½u m/s and constant velocity 2a m/s.

Given that when t=16, A is ahead of B, obtain in terms of u and a an expression for the distance between the particles at that time

Given also that this distance is 12m, and that the velocity of A when t=16 is 10m/s calculate

i - the value of u and of a

Hence calculate when t=18

ii- the distance between the particles
iii- the difference between their velocities

[Jimmy]

1)
i)

Consider X

using v=u+at
v=16+(-6)t
v=16-6t (1)

Consider Y

using v=u+at
v=8+2t (2)

equate (1) and (2)

16-6t = 8+2t
8=8t
t=1

when t=1 for Y
using s=ut+0.5at
s=8(1)+0.5(2)(1)
s=9m


ii)
For X
v=u+at
v=16+(-6)(1)
v=10 m/s

2)
a)i)
using v=u+2as
0=12+2(-9.8)h
0=144-19.6h
h=144/19.6
h=7.3469...
h=7.3m (2sf)

ii)
using v=u+2as
v=12+2(-9.8)(4)
v=65.6
v=8.099..
v=8.1m/s (2sf)

Hope his helps - have no time at present to do 3

[MeatLoafRocks]

2b)
Two Particles A and B are 96m apart on a smooth horizontal surface. A is moving directly towards B with speed 7m/s and retardation 2m/s. B is moving directly towards A with speed with 3m/s and accelerating 4m/s. Calculate the time taken before they meet


Answer
Consider A
let dist travelled by A = x
using s=ut+0.5at
x=7t+0.5(-2)t
x=7t-t

Consider B
let dist. travelled be y
using s=ut+0.5t
y=3t+0.5(4)t
y=3t+2t

but x+y = 96
so 7t-t+3t+2t=96
10t+t=96
t+10t-96=0
(t+16)(t-6)=0
t=-16 or t=6
since t>0 then t=6 seconds

[IsOLaTiOnIsT]

no worries, thanx for ur help, i got q1 and a bit of 2 but i just wanted to check it, i havent done 3 though, for 2ai and ii i got a different answer but then i forgot to state "take g to be 10m/s " leading to 'g=a' they give accurate answers without rounding, i got 7.2m and 8m/s, but yeh cheers for ur help, im just starting 3 now

[IsOLaTiOnIsT*]

I've done three, cheers for ur help :)