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?