Finding coordinates after finding the acceleration of an object

[nerdalert21]

Acceleration and coordinates at time t.

Homework Statement

At t = 0, a particle moving in the xy plane with constant acceleration has a velocity of vector v i = (3.00 i - 2.00 j) m/s and is at the origin. At t = 3.60 s, the particle's velocity is vector v = (7.00 i + 3.70 j) m/s. (Use the following as necessary: t. Round your coefficients to two decimal places.)

a) Find the acceleration of the particle at any time t. Found this a= (4.00i+5.70j)/3.6s
b) Find its coordinates at any time t. I have no clue how to do this

Homework Equations

I tried x⃗t=x⃗0+v⃗0t+12a⃗t^2 but webassign says my answer is wrong

The Attempt at a Solution

So i used the above equation and plugged in what I am given or know.
Xo=0
volt=(3.00i-2.00j)t
a=(4.00i+5.70j)/3.60
soooo
Xt= 0 + (3.00i-2.00j)t + 1/2[(4.00i+5.70j)/3.60]t^2
in the end i got
x=(3.00i)t + (0.56i)t^2
y=(2.00j)t + (0.79j)t^2

Can someone help me understand why i got this wrong?

 

[jacket]

a = 1.11 i + 1.58 j
Integrating,
v = 1.11 t i + 1.58 t j
Again, integrating,
s = 1.11 t$^{2}$/2 i + 1.58 t$^{2}$/2 j

 

[nerdalert21]

a = 1.11 i + 1.58 j
Integrating,
v = 1.11 t i + 1.58 t j
Again, integrating,
s = 1.11 t$^{2}$/2 i + 1.58 t$^{2}$/2 j

I understand where the a= 1.11i + 1.58j comes from but I don't understand what the other two parts (v and s) have to do with this
Can you explain?

 

[jacket]

s = 1.11 t$^{2}$/2 i + 1.58 t$^{2}$/2 j
so
x = 1.11 t$^{2}$/2
y = 1.58 t$^{2}$/2

 

[nerdalert21]

s = 1.11 t$^{2}$/2 i + 1.58 t$^{2}$/2 j
so
x = 1.11 t$^{2}$/2
y = 1.58 t$^{2}$/2

So your saying I have no need to use Xt= Xo + volt + 1/2at^2?
But what is s?
Cause it looks like your saying that by using the i and j components of "s", the x and y would be constants
but that's not possible is it? because I am looking for the coordinates at any time t

 

[jacket]

Unfortunately I never used proper limits in integration. Anyway let's do it with your equation.
By s I meant displacement vector.
Now,
$\overline{s}$ = $\overline{v0}$t + 1/2 $\overline{a}$t$^{2}$
$\Rightarrow$ $\overline{s}$ = (3i - 2j)t + 1/2 (1.11i + 1.58j)t$^{2}$
$\Rightarrow$ $\overline{s}$ = (3t +.56t^2)i + (-2t + .79t^2)j
Thus finally,
x = (3t + .56t^2)
y = (-2t + .79t^2)

 

[nerdalert21]

ohhh ok that makes sense
I almost had it right but you cleared up my confusion
thank you so much!

 

[jacket]

Correcting my first approach:

$\bar{a}$ = 1.11i + 1.58j
Integrating, $\bar{v}$ - $\bar{v0}$ = 1.11ti + 1.58tj
Again integrating, $\bar{s}$ - $\bar{s0}$- $\bar{v0}$t = 1.11 (t^2)/2 i + 1.58 (t^2)/2 j
But $\bar{s0}$ = 0 as the particle is at origin at t = 0
So, $\bar{s}$ = $\bar{v0}$t + 1.11 (t^2)/2 i + 1.58 (t^2)/2 j
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