## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

(a) The magnitude of the vector is 7.6 and the direction is an angle of $66.8^{\circ}$ above the positive x-axis. (b) The magnitude of the vector is $4.9~m/s^2$ and the direction is an angle of $66.0^{\circ}$ above the negative x-axis. (c) The magnitude of the vector is 17.8 m/s and the direction is an angle of $38.2^{\circ}$ below the positive x-axis. (d) The magnitude of the vector is $4.0~m$ and the direction is an angle of $56.3^{\circ}$ below the negative x-axis.
(a) We can find the magnitude of the vector. $A = \sqrt{A_x^2+A_y^2}$ $A = \sqrt{(3.0)^2+(7.0)^2}$ $A = 7.6$ We can find the angle above the positive x-axis. $tan(\theta) = \frac{7.0}{3.0}$ $\theta = tan^{-1}(\frac{7.0}{3.0}) = 66.8^{\circ}$ The magnitude of the vector is 7.6 and the direction is an angle of $66.8^{\circ}$ above the positive x-axis. (b) We can find the magnitude of the vector. $a = \sqrt{a_x^2+a_y^2}$ $a = \sqrt{(-2.0~m/s^2)^2+(4.5~m/s^2)^2}$ $a = 4.9~m/s^2$ We can find the angle above the negative x-axis. $tan(\theta) = \frac{4.5}{2.0}$ $\theta = tan^{-1}(\frac{4.5}{2.0}) = 66.0^{\circ}$ The magnitude of the vector is $4.9~m/s^2$ and the direction is an angle of $66.0^{\circ}$ above the negative x-axis. (c) We can find the magnitude of the vector. $v = \sqrt{v_x^2+v_y^2}$ $v = \sqrt{(14~m/s)^2+(-11~m/s)^2}$ $v = 17.8~m/s$ We can find the angle below the positive x-axis. $tan(\theta) = \frac{11}{14}$ $\theta = tan^{-1}(\frac{11}{14}) = 38.2^{\circ}$ The magnitude of the vector is 17.8 m/s and the direction is an angle of $38.2^{\circ}$ below the positive x-axis. (d) We can find the magnitude of the vector. $r = \sqrt{r_x^2+r_y^2}$ $r = \sqrt{(-2.2~m)^2+(-3.3~m)^2}$ $r = 4.0~m$ We can find the angle below the negative x-axis. $tan(\theta) = \frac{3.3}{2.2}$ $\theta = tan^{-1}(\frac{3.3}{2.2}) = 56.3^{\circ}$ The magnitude of the vector is $4.0~m$ and the direction is an angle of $56.3^{\circ}$ below the negative x-axis.