Finding coordinates after finding the acceleration of an object

1. Sep 15, 2013

Acceleration and coordinates at time t.

1. The problem statement, all variables and given/known data
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

2. Relevant equations

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

3. The attempt at a solution
So i used the above equation and plugged in what I am given or know.
Xo=0
Vot=(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?

Last edited: Sep 15, 2013
2. Sep 15, 2013

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

3. Sep 15, 2013

I understand where the a= 1.11i + 1.58j comes from but I dont understand what the other two parts (v and s) have to do with this
Can you explain?

4. Sep 15, 2013

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

5. Sep 15, 2013

So your saying I have no need to use Xt= Xo + Vot + 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 thats not possible is it? because im looking for the coordinates at any time t

6. Sep 15, 2013

jacket

Unfortunately I never used proper limits in integration. Anyway lets 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)

Last edited: Sep 15, 2013
7. Sep 15, 2013

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

8. Sep 15, 2013

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
...

Last edited: Sep 15, 2013