Answer
(a) The vector $a$ has a magnitude of $5.0~m/s^2$
(b) The vector $a$ is directed at an angle of $36.9^{\circ}$ counter-clockwise from the positive y-axis.
Work Step by Step
(a) We can find the magnitude of $a$:
$\vert a \vert = \sqrt{(-3.0~m/s^2)^2+(4.0~m/s^2)^2}$
$\vert a \vert = 5.0~m/s^2$
The vector $a$ has a magnitude of $5.0~m/s^2$
(b) We can find the angle $\theta$ counter-clockwise from the positive y-axis:
$tan(\theta) = \frac{3.0}{4.0}$
$\theta = arctan(\frac{3.0}{4.0})$
$\theta = 36.9^{\circ}$
The vector $a$ is directed at an angle of $36.9^{\circ}$ counter-clockwise from the positive y-axis.
