# 2D Kinematics Problem

1. Sep 22, 2010

### praecox

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

A moderate wind accelerates a pebble over a horizontal xy plane with a constant acceleration = (4.60 m/s2) + (7.00 m/s2). At time t = 0, the velocity is (4.4 m/s). What are magnitude and angle of its velocity when it has been displaced by 11.5 m parallel to the x axis?

2. Relevant equations

rf = (xi + Vxit + 1/2 axt2) i + (yi + Vyit + 1/2 ayt2) j
... I think

3. The attempt at a solution
I missed a day of class due to the flu, and our text book sucks at explaining. I think the variables should be:
ax = 4.6
ay = 7
vxi = 4.4
vyi = 0
xi = 0
yi = 0
xf = 11.5
So all I need to solve for is time, yes?
I tried to use a 1D equation to find t, since we have the Vi and xf.
I plugged the x values into:
xf - xi = Vxit + 1/2 axt2.
11.5 - 0 = 4.4t - 2.3t2
t2 + 1.91t + (0.91) = 5 + 0.91 (divided by 2.3 and then completed the square)
(t + 0.96)2 = 5.91
t + 0.96 = 2.43
t = 1.47 sec

When I plug in 1.47 as the time though, and then take the magnitude of the rf vector, I get 13.7 m, but my online homework says the answer is 15.2.

I've tried a few other equations, but I just can't seem to get it... and that's just the first half of the problem. I'm not even sure where to start to find the angle. :/
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 22, 2010

### Thaakisfox

It seems that you changed a sign. it should be 4.4t+2.3t^2 and you wrote -2.3t^2

3. Sep 22, 2010

### praecox

Oh, yes. That was just a typo. when I did the calculation, I did it with 11.5 - 0 = 4.4t + 2.3t2.