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

(a) The magnitude of the vector is 7.2 and the direction is an angle of $56.3^{\circ}$ below the positive x-axis. (b) The magnitude of the vector is 94.3 m and the direction is an angle of $58^{\circ}$ above the positive x-axis. (c) The magnitude of the vector is 44.7 m/s and the direction is an angle of $63.4^{\circ}$ above the negative x-axis. (d) The magnitude of the vector is $6.3~m/s^2$ and the direction is an angle of $71.6^{\circ}$ below the positive x-axis.
(a) We can find the magnitude of the vector. $A = \sqrt{A_x^2+A_y^2}$ $A = \sqrt{(4)^2+(-6)^2}$ $A = 7.2$ We can find the angle below the positive x-axis. $tan(\theta) = \frac{6}{4}$ $\theta = tan^{-1}(\frac{6}{4}) = 56.3^{\circ}$ The magnitude of the vector is 7.2 and the direction is an angle of $56.3^{\circ}$ below the positive x-axis. (b) We can find the magnitude of the vector. $r = \sqrt{r_x^2+r_y^2}$ $r = \sqrt{(50~m)^2+(80~m)^2}$ $r = 94.3~m$ We can find the angle above the positive x-axis. $tan(\theta) = \frac{80}{50}$ $\theta = tan^{-1}(\frac{80}{50}) = 58^{\circ}$ The magnitude of the vector is 94.3 m and the direction is an angle of $58^{\circ}$ above the positive x-axis. (c) We can find the magnitude of the vector. $v = \sqrt{v_x^2+v_y^2}$ $v = \sqrt{(-20~m/s)^2+(40~m/s)^2}$ $v = 44.7~m/s$ We can find the angle above the negative x-axis. $tan(\theta) = \frac{40}{20}$ $\theta = tan^{-1}(\frac{40}{20}) = 63.4^{\circ}$ The magnitude of the vector is 44.7 m/s and the direction is an angle of $63.4^{\circ}$ above the negative x-axis. (d) We can find the magnitude of the vector. $a = \sqrt{a_x^2+a_y^2}$ $a = \sqrt{(2.0~m/s^2)^2+(-6.0~m/s^2)^2}$ $a = 6.3~m/s^2$ We can find the angle below the positive x-axis. $tan(\theta) = \frac{6.0}{2.0}$ $\theta = tan^{-1}(\frac{6.0}{2.0}) = 71.6^{\circ}$ The magnitude of the vector is $6.3~m/s^2$ and the direction is an angle of $71.6^{\circ}$ below the positive x-axis.