Direction Tests

Concept:Find net displacement by resolving all movements into east-west and north-south components.Explanation:Assume all distances in meters. Let east be positive $x$, north be positive $y$.First: $7$ m SE at $45^\circ$ south of east: $dx_1 = 7 \cos45^\circ = \frac{7}{\sqrt{2}}$ m, $dy_1 = -7 \sin45^\circ = -\frac{7}{\sqrt{2}}$ m.Second: $14$ m West: $dx_2 = -14$ m, $dy_2 = 0$.Third: $7$ m NW at $45^\circ$ north of west: $dx_3 = 7 \cos135^\circ = -\frac{7}{\sqrt{2}}$ m, $dy_3 = 7 \sin135^\circ = \frac{7}{\sqrt{2}}$ m.Fourth: $4$ m East: $dx_4 = 4$ m, $dy_4 = 0$.Total $dx = \frac{7}{\sqrt{2}} - 14 - \frac{7}{\sqrt{2}} + 4 = -10$ m.Total $dy = -\frac{7}{\sqrt{2}} + 0 + \frac{7}{\sqrt{2}} + 0 = 0$ m.Net displacement is $10$ m west, so distance from start is $10$ m.Answer:$10$ m