Question

1. A hiker begins her trip away from her car by first walking 30.0 km west. She stops and sets up her tent for the night. On the second day, she walks 20.0 km in a direction that makes an angle of 35.0 north of west, at which point she discovers a forest ranger's tower. Determine the magnitude of her total displacement km 2. A vector A has a magnitude of 50.0 m and points in a direction 28.0° above the negative x-axis. A second vector, B, has a magnitude of 80.0 m and points in a direction 60.0 above the negative x-axis. Using the component method, find the magnitude of the vector R- Ä+ B. 3. A vector Ä has a magnitude of 50.0 m and points in a direction 28.0° above the negative x-axis. A second vector, B, has a magnitude of 80.0 m and points in a direction 60.0 above the negative x-axis. Using the component method, find the magnitude of the vector D- A - B. 4 Consider the three displacement vectors shown in the figure: Vector A has a magnitude of 7.50 km and a direction that makes an angle θ 290% to the left of the positive y axis vector has a magnitude of 5 80 km and a direction that makes an angle of α 35.00 above the positive x-axis, and vector C has a magnitude of 3.10 km and a direction that makes an angle B 67.0° below the negative x-axis. Determine the magnitude of the vector D- A+B- km

Answer 1

Total displacement (OB^vector) = OA^vector + AB^vector OA^vector = -30 i^^km AB = -20 cos 35 i^^+ 20 sin 35 j^^= -16.38 i^^+ 11.47 j^^km OA^vector + AB^vector = -46.38 i^^+ 11.47 j^^Displacement = squareroot 46.38^2 + 11.41^2 D^vector = 47.78 km