Question

QUESTION 3 : Relative motion and 2d vectors

Question 3 (relative motion & 2d vectors) There are two ports, A and B. The displacement from port A to port B is 1600 m at 40% (or 40% north of east). You have a boat that travels 400 meters per minute in the water. Further, there is a constant current in the water that flows 80 meters per minute at 110% (or 20° west of north) as observed from the shore. You travel from port A to port B in a straight line. Here are diagrams for both the positions and velocities. positions in red velocities in black port B Vboat by shore v @ 40° Vwater by shore 80 m/min @ 110 0° port A a. Sketch the approximate vector that is the velocities of the boat as observed by the water. b. Write the equation that describes the velocity of the boat as observed by the water in terms of the other velocities above? c. What is the speed of the boat as observed by the shore (v in the diagram above)? d. At what angle is the boat traveling as observed by the water (angle of your sketched velocity vector)?

Answer 1

V_boat by water = V_boat by - V_water by c In x-y co-ordinate (400 cos theta i + 400 sin theta j) = (- 80 sin 20i + 80 cos 20j) + (V cos 40i + V sin 40j) Along x-axis 400 cos theta = V cos 40 - 80 sin 20 A Along Y-axis 400 sin theta = V sin 40 degree + 80 cos 20 B from A and B (400)^2 = (V cos 40 - 80 sin 20)^2 + (V sin 40 + 80 cos 20)^2 (400)^2 = V^2 + 80^2 - 160V cos 40 degree sin 20 degree + 160V sin 40 cos 20 degree implies V^2 + 160V sin(40 - 20) - 153600 = 0 V = -160 sin 20 plusminus squareroot (160 sin 20)^2 + 4 times 1 times 153600/2 times 1 m/min implies V = 365.5 m/min d from equation A 400 cos theta = 365.5 times cos 40 degree - 80 times sin 20 degree implies cos theta = 0.6316