Physics: Principles with Applications (7th Edition)

a. 56$^o$ b. 140s
a. Let the flow of the river be in the x direction and across the river be the y direction. Let S denote Swimmer, R the River, and B the Bank. The pair “SB”, for example, represents the swimmer’s motion relative to the bank. $$\vec{v_{SB}} = \vec{v_{SR}} + \vec{v_{RB}}$$ $$(0, V) = (0.60 cos \theta, 0.60 sin \theta) m/s + (0.50, 0) m/s$$ Equate the x-components to solve for the angle of 146.44$^o$, or 56$^o$ west of north. b. Find the crossing time from the y-direction. First find the velocity in the y-direction. $$(0, V) = (0.60 cos \theta, 0.60 sin \theta) m/s + (0.50, 0) m/s$$ $$V = (0.60 m/s)sin 146.44 ^{\circ} + 0 = 0.332 m/s$$ The time to cross is t = (45 m)/(0.332m/s) =135.68s $\approx$ 140s