## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

(a) The magnitude of the vector is 5.7 and the direction is an angle of $45^{\circ}$ above the negative x-axis. (b) The magnitude of the vector is 2.2 cm and the direction is an angle of $26.6^{\circ}$ below the negative x-axis. (c) The magnitude of the vector is 100.5 m/s and the direction is an angle of $84^{\circ}$ below the negative x-axis. (d) The magnitude of the vector is $22~m/s^2$ and the direction is an angle of $26.6^{\circ}$ above the positive x-axis.
(a) We can find the magnitude of the vector. $B = \sqrt{B_x^2+B_y^2}$ $B = \sqrt{(-4)^2+(4)^2}$ $B = 5.7$ We can find the angle above the negative x-axis. $tan(\theta) = \frac{4}{4}$ $\theta = tan^{-1}(\frac{4}{4}) = 45^{\circ}$ The magnitude of the vector is 5.7 and the direction is an angle of $45^{\circ}$ above the negative x-axis. (b) We can find the magnitude of the vector. $r = \sqrt{r_x^2+r_y^2}$ $r = \sqrt{(-2.0~cm)^2+(-1.0~cm)^2}$ $r = 2.2~cm$ We can find the angle below the negative x-axis. $tan(\theta) = \frac{1.0}{2.0}$ $\theta = tan^{-1}(\frac{1.0}{2.0}) = 26.6^{\circ}$ The magnitude of the vector is 2.2 cm and the direction is an angle of $26.6^{\circ}$ below the negative x-axis. (c) We can find the magnitude of the vector. $v = \sqrt{v_x^2+v_y^2}$ $v = \sqrt{(-10~m/s)^2+(-100~m/s)^2}$ $v = 100.5~m/s$ We can find the angle below the negative x-axis. $tan(\theta) = \frac{100}{10}$ $\theta = tan^{-1}(\frac{100}{10}) = 84^{\circ}$ The magnitude of the vector is 100.5 m/s and the direction is an angle of $84^{\circ}$ below the negative x-axis. (d) We can find the magnitude of the vector. $a = \sqrt{a_x^2+a_y^2}$ $a = \sqrt{(20~m/s^2)^2+(10~m/s^2)^2}$ $a = 22~m/s^2$ We can find the angle above the positive x-axis. $tan(\theta) = \frac{10}{20}$ $\theta = tan^{-1}(\frac{10}{20}) = 26.6^{\circ}$ The magnitude of the vector is $22~m/s^2$ and the direction is an angle of $26.6^{\circ}$ above the positive x-axis.