A woman who can row a boat at 6.4 km/h in still water faces a long, straight river with a width of 6.3 km and a current of 3.3 km/h. Let i point directly across the river and j point directly downstream. If she rows in a straight line to a point directly opposite her starting position, (a) at what angle to i must she point the boat (b) how long (in hours) will she take? (c) How long (in hours) will she take if, instead, she rows 3.6 km down the river and then back to her starting point? (d) How long (in hours) if she rows 3.6 km up the river and then back to her starting point? (e) At what angle to i should she point the boat if she wants to cross the river in the shortest possible time? (f) How long (in hours) is that shortest time? In your response, please clearly indicate which portion of the question you are answering, and detail how you reached that answer. Additionally, in this problem i = x-axis and j = y-axis, though the i (x) is the vertical axis and the j (y) is the horizontal axis in this problem 2. Pythagorean theorem The y component of the angle in part a is cos (theta) = y/(6.4 km/h) The x component of the angle in part a is sin (theta) Theta in part a is arctan(x component of vector/y componenet of vector) Velocity*time = displacement A.) , I know the x and y components of the vector 6.4 (km/h) needs to be found in order to use arctan(x/y) to find the angle theta that the boat makes with the x (vertical/i) axis, and that the width of the river might be used somehow to find the x(i) vector component) B.) For part b, the velocity of the boat (Vb (wrt w) with respect to the water needs to be found using the velocity of the boat with respect to the ground (Vb wrt g) and the velocity of the boat with respect to the ground (VB wrt g). I tried (VB wrt g) - (VW wrt g) = (VB wrt w) and received the answer 3.1 (km/h), but when this is multiplied by the 6.3 km she needs to travel, it results that it will take her 19.53 hours to make the crossing, but this does not follow common logic, and also starkly deviates from the answer to a problem similar to this (same scenario, different numerical values).