91Ó°ÊÓ

To find Sheena's total velocity with respect to the bank, we will break her boat's velocity into horizontal and vertical components in relation to the width of the river and the given angle. We have: Horizontal component: \(v_{bx} = v_b \cos(\theta) = 3.00 \mathrm{mi/h} \cos(60.0^\circ) = 1.50\mathrm{mi/h}\) Vertical component: \(v_{by} = v_b \sin(\theta) = 3.00 \mathrm{mi/h} \sin(60.0^\circ) = 2.60\mathrm{mi/h}\) Next, we will calculate her total velocity in both the horizontal and vertical directions by adding the river current in the horizontal direction: Total horizontal velocity: \(v_{tx} = v_{bx} + v_{r} = 1.50\mathrm{mi/h} + 1.60\mathrm{mi/h} = 3.10\mathrm{mi/h}\) Total vertical velocity: \(v_{ty} = v_{by} = 2.60\mathrm{mi/h}\) Using the Pythagorean theorem, we can find her resultant total velocity: \(v_t = \sqrt {v_{tx}^2 + v_{ty}^2} = \sqrt{(3.10\mathrm{mi/h})^2+(2.60\mathrm{mi/h})^2} = 4.031\mathrm{mi/h}\) Hence, Sheena's speed with respect to the starting point on the bank is \(4.031\mathrm{mi} / \mathrm{h}\)

To find the time it takes Sheena to cross the river, we can use the distance and her vertical velocity, since this is the velocity component that takes her across the river. Time = Distance / Vertical Velocity = Width of the river / \(v_{ty}\) \(t = \frac{1.20\mathrm{mi}}{2.60\mathrm{mi/h}}= 0.4615\mathrm{h}\) Sheena takes 0.4615 hours or 27.69 minutes to cross the river.

To find how far upstream or downstream Sheena will be from her starting point, we will use her total horizontal velocity, which accounts for both her boat's speed and the river current, and the time calculated to reach the other bank. Horizontal Displacement = Total Horizontal Velocity × Time \(D_x = v_{tx} \times t = 3.10\mathrm{mi/h} \times 0.4615\mathrm{h} = 1.43\mathrm{mi}\) She will be 1.43 miles downstream from her starting point when she reaches the opposite bank of the river.

To determine the correct angle upstream Sheena should've aimed for, we can consider the velocity components required for her to go straight across the river. We know that the vertical component of her boat’s velocity should be greater than or equal to her boat's speed (\(v_{b}=3.00\mathrm{mi/h}\)) to counteract the river current in the horizontal direction. To achieve this, we keep the horizontal component equal to the river current: \(v_{bx} = v_r = 1.60\mathrm{mi/h}\) Now, we can find the required angle utilizing the boat’s velocity: \(\cos(\alpha) = \frac{v_{bx}}{v_{b}} = \frac{1.60\mathrm{mi/h}}{3.00\mathrm{mi/h}} = 0.5333\) \(\alpha = \arccos(0.5333) = 57.91^{\circ}\) To go straight across the river, Sheena should have aimed for an upstream angle of \(57.91^{\circ}\).