Acceleration is (-2.0i + 4.0j), velocity is 12 m/s

[new^2^physics]

Homework Statement

At T=0 a particle leaves the origin with a velocity of 12 m/s in the positive x direction and moves in the xy plane with a constant acceleration of (-2.0i + 4.0j)m/s^2. At the instant the y coordinate of the particle is 18 m, what is the x coordinate of the particle.

Homework Equations

rf=ri + vit + 1/2at²
=(12m/s)(t) + 1/2 (-2.0i + 4.0j)t²

So stuck. Any direction or help would be greatly appreciated!

 

[The legend]

so you know initial velocity to be 12.0i..
now you also know a = dv/dt
and v = dr/dt (r being position)..

Which mathematical (or calculus) tool can you use to get your answer now? and how?

 

[new^2^physics]

Okay, so I am going to play with these equations and try to get closer...

a = dv/dt
(-2.0i + 4.0j)= 12i / dt

dt = 12i / -6.0i + 4.0j
dt = -6.0i - 4.0 j

v = dr/dt
12 i = xi - 18j / -6.0i - 4.0j
12i(-6.0i - 4.0 j) = xi - 18j

oh dear, now I am stuck again and I am not even sure I have done this part correctly.

 

[The legend]

Okay, so I am going to play with these equations and try to get closer...

a = dv/dt
(-2.0i + 4.0j)= 12i / dt

that isn't correct...

a=dv/dt
(-2.0i+4.0j)dt=dv
integrating

-(2t)i + (4t)j = v-12

so you get v as a function of time... now proceed in a similar manner...

 

[zgozvrm]

At T=0 a particle leaves the origin with a velocity of 12 m/s in the positive x direction...

So, what are V_{ix}[/tex] and V_{iy}[/tex]?