Answer
(a) $v_{av,x} = 1.4~m/s$
$v_{av,y} = -1.3~m/s$
(b) $v_{av} = 1.9~m/s$
The angle below the positive x-axis is $42.9^{\circ}$
Work Step by Step
At $t_1 = 0$, the coordinates of the squirrel are $(1.1~m, 3.4~m)$.
At $t_2 = 3.0~s$, the coordinates of the squirrel are $(5.3~m, -0.5~m)$.
(a) $v_{av,x} = \frac{\Delta x}{\Delta t} = \frac{5.3~m-1.1~m}{3.0~s} = 1.4~m/s$
$v_{av,y} = \frac{\Delta y}{\Delta t} = \frac{-0.5~m-3.4~m}{3.0~s} = -1.3~m/s$
(b) We can find the magnitude of the average velocity $v_{av}$.
$v_{av} = \sqrt{(v_{av,x})^2+(v_{av,y})^2}$
$v_{av} = \sqrt{(1.4~m/s)^2+(-1.3~m/s)^2}$
$v_{av} = 1.9~m/s$
We can find the angle below the positive x-axis.
$tan(\theta) = \frac{1.3}{1.4}$
$\theta = tan^{-1}(\frac{1.3}{1.4}) = 42.9^{\circ}$
