#### Answer

(a) The puck is moving at an angle of $61.9^{\circ}$ above the positive x-axis.
(b) The puck is a distance of 180 cm from the origin.

#### Work Step by Step

(a) The acceleration in the x-direction is the slope of the $v_x$ versus time graph;
$a_x = \frac{40~cm/s}{5~s} = 8~cm/s^2$
At t = 2 s:
$v_x = a_x~t = (8~cm/s^2)(2~s)$
$v_x = 16~cm/s$
$v_y = 30~cm/s$
We can find the angle $\theta$ above the positive x-axis;
$tan(\theta) = \frac{30~cm/s}{16~cm/s}$
$\theta = arctan(\frac{30~cm/s}{16~cm/s})$
$\theta = 61.9^{\circ}$
The puck is moving at an angle of $61.9^{\circ}$ above the positive x-axis.
(b) The displacement in the x-direction is the area under the $v_x$ versus time graph.
$x = \frac{1}{2}(40~cm/s)(5~s)$
$x = 100~cm$
The displacement in the y-direction is the area under the $v_y$ versus time graph.
$y = v_y~t$
$y = (30~cm/s)(5~s)$
$y = 150~cm$
We can find the distance $d$ from the origin.
$d = \sqrt{(x)^2+(y)^2}$
$d = \sqrt{(100~cm)^2+(150~cm)^2}$
$d = 180~cm$
The puck is a distance of 180 cm from the origin.