A boat is heading due north as it crosses a wide river with a speed of 13.0 km/h

Question

A boat is heading due north as it crosses a wide river with a speed of 13.0 km/h relative to the water. The river has a current such that the water moves with a uniform speed 5.60 km/h due east relative to the ground.

(B) If the boat travels with the same speed of 13.0 km/h relative to the river and is to travel due north as shown in figure (b), what should its heading be?

Conceptualize/Categorize This question is an extension of part (A), so we have already conceptualized and categorized the problem. One new feature of the conceptualization is that we must now aim the boat upstream so as to go straight across the river.

Analyze The analysis now involves the new triangle shown in figure (b). As in part (A), we know VrE and the magnitude of the vector br, and we want VbE to be directed across the river. Notice the difference between the triangle in figure (a) and the one in figure (b): the hypotenuse in figure (b) is no longer VbE.
Use the Pythagorean theorem to find bE:
VbE = (Vbr2 - VrE2)1/2 = km/h?
Find the direction in which the boat is heading:
? = tan-1(VrE / VbE) =

Explanation / Answer

b)Vbe+(-Vre)=Vbr fro m the figure Vbr sin?=Vre ?=25.51 Vbe=Vbrcos25.51=11.732km/h

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