## College Physics (4th Edition)

(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.
(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.