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**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

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?

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