A boat travels at \( 15 \mathrm{~m} / \mathrm{s} \) on still water. In what direction must the boat travel in order to reach its destination directly across the river? The river is moving \( 5.0 \mathrm{~m} / \mathrm{s} \). \( 90^{\circ} \) from the shore. \( 72^{\circ} \) from the shore downstream. \( 71^{\circ} \) from the shore upstream. \( 19^{\circ} \) from the shore upstream.

To determine the direction in which the boat must travel to reach its destination directly across the river, we can use vector analysis. Let: - \( v_b = 15 \, \text{m/s} \) (the speed of the boat in still water) - \( v_r = 5 \, \text{m/s} \) (the speed of the river current) We want the resultant velocity of the boat to be directly across the river (perpendicular to the current). 1. **Set up the vectors**: - The velocity of the river \( \vec{v_r} \) is directed downstream (let's assume to the right). - The velocity of the boat \( \vec{v_b} \) will have two components: one directly across the river (let's call this \( v_{b_y} \)) and one against the current (let's call this \( v_{b_x} \)). 2. **Components of the boat's velocity**: - The component of the boat's velocity across the river (perpendicular to the current) is \( v_{b_y} = v_b \sin(\theta) \). - The component of the boat's velocity against the current is \( v_{b_x} = v_b \cos(\theta) \). 3. **Setting up the equation**: To reach directly across the river, the upstream component of the boat's velocity must equal the downstream current: \[ v_{b_x} = v_r \] Therefore: \[ v_b \cos(\theta) = v_r \] Substituting the known values: \[ 15 \cos(\theta) = 5 \] \[ \cos(\theta) = \frac{5}{15} = \frac{1}{3} \] 4. **Finding \( \theta \)**: Now we can find \( \theta \): \[ \theta = \cos^{-1}\left(\frac{1}{3}\right) \] Using a calculator: \[ \theta \approx 72.34^\circ \] Since the boat must travel upstream to counteract the current, the angle \( \theta \) is measured from the shore upstream. Thus, the direction the boat must travel in order to reach its destination directly across the river is approximately \( 72^\circ \) from the shore upstream. **Final Answer**: \( 72^{\circ} \) from the shore upstream.