Part A A 400.0-I-wide river flows from west to east at 30.0 m/min. Your boat moves at 100.0 m/min relative to the water no matter which direction you point it. To cross this river, you start from a dock at point A on the south bank. There is a boat landing directly opposite at point B on the north bank, and also one at point C, 75.0 m downstream from B (the figure (Figure 1)). Where on the north shore will you land if you point your boat perpendicular to the water current? Submit Part B What distance will you have traveled, if you point your boat perpear to the water current? That is, the direction of your boat will remain perpendicular to the direction of the water current for the entire crossing. In Submit PartC If you initially aim your boat directly toward point C and do not change that bearing relative to the shore, where on the north shore will you land? Note: the wordling may cause some confusion. "initially aim" means that you point the boat in the direction of point C just as you leave the shore, and do not change the direction with respect to the water for the entire trip. In other words, once you leave shore, you only look don at the water and make sure that the direction vector does not change with respect to the water. You do NOT look at features on shore to make comections in the direction during transit so that you end up at C Im downstream from B Submit Part D To reach point C at what bearing (anglenorth of west) must you aim your boat? "Bearing" means the direction of the boat with respect to the water, not the land. Again, you should imagine that you can only look at the water to determine your direction (and not use a compass, or look at features on shore to make corrections in the direction) and that this direction is maintained for the entire trip across the river Hint he main point o his part o he problem s o encourage you o use the Galilean velocity vector o mula Let G be the ame attached to he ground, and Frame R be float ng with he ver. The object ofinterest hat s moving is he boat-label b. The velocity or mula becomes V G=VbR+VRG The issue here is that you only know the magnitude of the vector VbR (not the direction), and the direction of VbG but not the magnitude. The direction of VbG is given by the direction of vector that points from point A to point C. But, you have all the vector information for the direction of the water in the river, VRG. For these types of vector problems, you want to invoke the law of sines or the law of cosines because the 3 vectors in the Galilean velocity formula form a triangle. In this problem you have the lengths of two of the sides of the triangle and one angle. This is sufficient to get all the sides and all the angles using the law of sines and/or law of cosines north of west Figure 1 of 1 Submit Part Refer to part D. How long (t) will it take to cross the river? 4000 m Hint: Once you have the velocity vector, VbG (both the magnitude and direction) from the previous part D, find the component y" component (or the North-South component). The time is just the width of the river divided by the y component of velocity. min