#### Answer

At t = 5 s, the magnitude of the acceleration is $2.2~m/s^2$

#### Work Step by Step

The acceleration is the slope of the velocity versus time graph.
We can find $a_x$:
$a_x = \frac{\Delta v_x}{\Delta t}$
$a_x = \frac{20~m/s}{10~s}$
$a_x = 2.0~m/s^2$
We can find $a_y$:
$a_y = \frac{\Delta v_y}{\Delta t}$
$a_y = \frac{10~m/s}{10~s}$
$a_y = 1.0~m/s^2$
We can find the magnitude of the puck's acceleration;
$a = \sqrt{(a_x)^2+(a_y)^2}$
$a = \sqrt{(2.0~m/s^2)^2+(1.0~m/s^2)^2}$
$a = 2.2~m/s^2$
The magnitude of the acceleration is constant and it is equal to $2.2~m/s^2$. At t = 5 s, the magnitude of the acceleration is $2.2~m/s^2$.