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~christina~

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## 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

[tex]x(t)= A cos (\omega t + \phi )[/tex]

[tex]v(t)= - \omega A sin (\omega t + \phi ) [/tex]

[tex]a(t)= - \omega ^2 A cos (\omega t + \phi) [/tex]

## The Attempt at a Solution

well I know for the first situation that:

[tex]x_o[/tex]= 0.270

[tex]v_o[/tex]= 0.140m/s

[tex]a_o[/tex]= -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

[tex]v(t)= - \omega A sin (\omega t + \phi ) [/tex]

though I need the [tex] \phi [/tex]...hm..

and I think I'm lost here.

can someone Please help me out with this?

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