## University Physics with Modern Physics (14th Edition)

(a) Magnitude: 9.01 m; Direction: $33.7^o$ (b) Magnitude: 9.01 m; Direction: $33.7^o$ (c) Magnitude: 22.3 m; Direction: $250^o$ (d) Magnitude: 22.3 m; Direction: $70.3^o$
As we have calculated in the exercise 1.27: $$A_x = 0 , A_y = -8.00 \space m$$ $$B_x = 7.50 \space m, B_y = 13.0 \space m$$ (a) If $\vec R = \vec A + \vec B$: $$R_x = A_x + B_x = 0 + 7.50 \space m = 7.50 \space m$$ $$R_y = A_y + B_y = -8.00 \space m + 13.0 \space m = 5.00 \space m$$ $$R = \sqrt{R_x^2 + R_y^2} = \sqrt{7.50^2 + 5.00^2}\space m = 9.01 \space m$$ $$\theta = arctan \space \frac{R_y}{R_x} = arctan \space \frac{5.00}{7.50} =33.7 ^o$$ (b) If $\vec R = \vec B + \vec A$: $$R_x = B_x + A_x = 7.50 \space m + 0= 7.50 \space m$$ $$R_y = B_y + A_y = 13.0 \space m -8.00 \space m = 5.00 \space m$$ $$R = \sqrt{7.50^2 + 5.00^2}\space m = 9.01 \space m$$ $$\theta = arctan \space \frac{5.00}{7.50} =33.7 ^o$$ (c) If $\vec R = \vec A - \vec B$: $$R_x = A_x - B_x = 0 - 7.50 \space m = -7.50 \space m$$ $$R_y = A_y - B_y = - 8.00 \space m - 13.0 \space m = -21.00 \space m$$ $$R = \sqrt{(-7.50)^2 + (-21.00)^2}\space m = 22.3 \space m$$ $$\theta = arctan \space \frac{-21.00}{-7.50} = 70.3 ^o$$ But, since $R_x \lt 0$ and $R_y \lt 0$, the angle must be in the third quadrant. Since that quadrant starts at $180^o$: $$\theta = 70.3^o + 180^o = 250^o$$ (d) If $\vec R = \vec B - \vec A$: $$R_x = B_x - A_x = 7.50 \space m - 0= 7.50 \space m$$ $$R_y = B_y - A_y = 13.0 \space m -(-8.00) \space m = 21.0 \space m$$ $$R = \sqrt{7.50^2 + (21.0)^2}\space m = 22.3 \space m$$ $$\theta = arctan \space \frac{21.0}{7.50} =70.3 ^o$$