Simple harmonic motion of a particle

[~christina~]

Homework Statement

A particle moves along x-axis It is initially at 0.270m moving with velocity of 0.140m/s
and acceleration of -0.320m/s^2
Suppose it moves with constant acceleration for 4.50s.

Find it's:
a) position and velocity at end of time interval

next assume it moves with simple harmonic motin for 4.50s and x= 0 is it's equillibrium position.

Find it's:

b) position and velocity at end of this time interval

Homework Equations

x(t)= A cos (\omega t + \phi )
v(t)= - \omega A sin (\omega t + \phi )
a(t)= - \omega ^2 A cos (\omega t + \phi)

The Attempt at a Solution

well I know for the first situation that:

x_o= 0.270
v_o= 0.140m/s
a_o= -0.320m/s^2

and the acceration remain constant for 4.50s

to find a) the postiion and velocity at end of time interval...

I was thinking that I would take the period (T) to equal T= 4.50s?
and would I use 4.51 seconds and use the velocity as a function of time equation

v(t)= - \omega A sin (\omega t + \phi )

though I need the \phi...hm..

and I think I'm lost here.

can someone Please help me out with this?

 

[Littlepig]

so, a) it's no problem as is an 1 dimension problem, with movement equations...

about b), it's a whole new problem:
you will need to know the position and velocity at t_0=4.50
so, you can write: t_1=0 at x_1=0(equilibrium point)
by a), you calculated velocity at t_0, so, it's the particles velocity at t_1

I guess this should help...