How can I find the magnitude of the acceleration in this $2$-D motion kinematic equation? Can someone help me plug in the correct variables for this $2$-D motion problem?
So the problem is: A puck moves $2.35 \frac{m}{s}$ in a $-22.0^\circ$ direction. A hockey stick pushes it for $0.215$ s, changing its velocity to $6.42 \frac{m}{s}$ in a $50.0^\circ$ direction. What was the magnitude of the acceleration?
Here I have the given vectors for the X-component:
$$v_{ix} (velocity initial) = 2.35\cos(-22.0) = 2.17$$
$$v_{fx} (velocity final) = 6.42\cos(50.0) = 4.12$$
$$a_x (acceleration) = missing$$
$$\Delta x = missing$$
$$t (time) = 0.215s$$
And the given vectors for the Y-component:
$$v_{iy} = 2.35\sin(-22.0) = -.880$$
$$v_{fy} = 6.42\sin(50.0) = 4.91$$
$$a_y = unknown$$
$$\Delta y = missing$$
$$t = 0.215s$$
The formula for when you are missing $\Delta x$ is $v_f = v_i+at$, which I rearranged to $(v_f-v_t)/t = a$. 
For the X-component:
$4.12 - 2.17/0.215 = -5.97$
For the Y-component:
$4.91-(-.880)/0.215 = 11.95$
Isn't the formula to find magnitude $c = \sqrt{c_{x}^2 + c_y^2}$? So what variables would I use for $c_x$ and $c_y$? Or, is there a different formula for finding the magnitude of the acceleration? Or, where am I going wrong?
 A: The picture is approximate.
Use velocity and acceleration vectors projections to $x$ and $y$ axis. Angles are to $x$-axis
$v_2\cos\beta=v_1\cos\alpha+a_xt\ (1)$
$v_2\sin\beta=v_1\sin\alpha+a_yt\ (2)$
$a^2 = a_x^2 + a_y^2\ (3)$
Get $a_x$ and $a_y$ from (1) and (2) and put them to (3)
$a^2 = (\frac{v_2\cos\beta-v_1\cos\alpha}{t})^2 + (\frac{v_2\sin\beta-v_1\sin\alpha}{t})^2\ (4)$
or
$a = \frac{1}{t}\sqrt{(v_2\cos\beta-v_1\cos\alpha)^2 + (v_2\sin\beta-v_1\sin\alpha)^2}$
or
$a = \frac{1}{t}\sqrt{v_1^2+v_2^2 -2v_1v_2(\cos\beta\cos\alpha + \sin\beta\sin\alpha)}\ (4')$
As per @Andrei suggestion
$\cos\beta\cos\alpha + \sin\beta\sin\alpha = \cos(\beta-\alpha)$ see on Wikipedia
(4') goes to 
$ a = \frac{1}{t}\sqrt{v_1^2+v_2^2 -2v_1v_2\cos(\beta-\alpha)}\ (4'') $
Add numeric values to $(4'')$ and result will be $a=28.45 \frac{m}{s^2}$
Questions, errors, comments.
A: Using the Law of Cosine, you can find that the magnitude of the change in velocity is 
$$|\Delta v| = \sqrt{(2.35)^2+(6.42)^2-2(2.35)(6.42)\cos(50-(-22))} = 6.12\frac{m}{s}$$
$$|a| = \frac{|\Delta v|}{t} = \frac{6.12\frac{m}{s}}{0.215s} = 28.4 \frac{m}{s^2}$$
