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Chapter 3 Even Problem Solutions #12. The magnitude of the vector is: ||34î + 13ĵ|| = p 342 + 132 = 36.4m The angle it makes with the x-axis is: θ = tan−1 ( #28. The x component is: Vx = 18 The y m m ∗ cos(220◦ ) = −13.8 s s component is: Vy = 18 #48. 13 ) = 21◦ 34 m m ∗ sin(220◦ ) = −11.6 s s Let us make a few denitions. Let us dene the river ows, and x y to be the direction in which to be the directly across direction. Therefore, the ~vr = .57 m s ĵ . We will row, with respect to the shores, with a velocity ~ v = vx î + vy ĵ . Therefore, our total velocity is: ~vT = ~v + v~r = vx î + (vy + .57 m s )ĵ . river ows with velocity a. We want our total velocity to be 0 in the y direction. This is the denition of rowing straight across. Therefore, we nd immediately that: vy = −.57 m s We also know that we can row with total speed of 1.3 1.3 m s , therefore: p m q 2 = vx + vy2 = vx2 + (.57)2 s Solving this we get that: vx = 1.17 Because our y m s direction velocity is negative, we know we want to go against the ow of the river, or in the −y the direction we want to go as the angle θ = tan−1 ( b. t= 1 the x axis: .57 ) = 26◦ 1.17 We just noted that our (total) velocity in the therefore: direction. We can dene below 63m = 54s 1.17 m s x direction is vx = 1.17 m s ,