River crossing: reach point directly opposite (find angle and/or time)

4 A small ferry is used to cross a river which has straight parallel banks that are 200 metres apart. The water in the river moves at a constant speed of \(3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The ferry travels from a point \(A\) on one bank to a point \(B\) directly opposite \(A\) on the other bank. The velocity of the ferry relative to the water is \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of \(60 ^ { \circ }\) to the upstream bank, as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{d42b2e88-74ea-486b-bb47-f512eb0c185d-4_467_1122_550_459}

1. \(\quad\) Find \(V\).
2. Find the time that it takes for the ferry to cross the river, giving your answer to the nearest second.

6 A river has straight parallel banks. The water in the river is flowing at a constant velocity of \(3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) parallel to the banks. A boat crosses the river, from the point \(A\) to the point \(B\), so that its path is at an angle \(\alpha\) to the bank. The velocity of the boat relative to the water is \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) perpendicular to the bank. The diagram shows these velocities and the path of the boat. \includegraphics[max width=\textwidth, alt={}, center]{ccc1db66-9700-4f22-905e-cc0bdf1fd3c1-12_467_988_568_532}

1. Show that \(\alpha = 53.1 ^ { \circ }\), correct to three significant figures.
2. The boat returns along the same straight path from \(B\) to \(A\). Given that the speed of the boat relative to the water is still \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), find the magnitude of the resultant velocity of the boat on the return journey.

3 A boat can travel at a speed of \(2 \mathrm {~ms} ^ { - 1 }\) in still water. The boat is to cross a river in which a current flows at a speed of \(0.8 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The angle between the direction in which the boat is pointing and the bank is \(\alpha\). The boat travels so that the resultant velocity of the boat is perpendicular to the bank. \includegraphics[max width=\textwidth, alt={}, center]{7e0585ea-062a-487c-8e39-37a4ed414ff8-3_264_1040_575_493}

1. Show that \(\alpha = 66.4 ^ { \circ }\) correct to three significant figures.
2. 1. Find the magnitude of the resultant velocity of the boat.
   2. The width of the river is 14 metres. Find the time that it takes for the boat to cross the river.

4 A river has parallel banks which are 16 metres apart. The water in the river flows at \(1.2 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) parallel to the banks. A boat sets off from one bank at the point \(A\) and travels perpendicular to the bank so that it reaches the point \(B\), which is directly opposite the point \(A\). It takes the boat 10 seconds to cross the river. The velocity of the boat relative to the water has magnitude \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and is at an angle \(\alpha\) to the bank, as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{c022c936-72bc-4cf9-8f98-285f12c1d479-08_400_1011_667_511}

1. Show that the magnitude of the resultant velocity of the boat is \(1.6 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
2. \(\quad\) Find \(V\).
3. Find \(\alpha\).
4. State one modelling assumption that you needed to make about the boat.

   	\includegraphics[max width=\textwidth, alt={}]{c022c936-72bc-4cf9-8f98-285f12c1d479-08_72_1689_1617_154}

   .......... \(\_\_\_\_\) \includegraphics[max width=\textwidth, alt={}, center]{c022c936-72bc-4cf9-8f98-285f12c1d479-09_40_118_529_159}

\includegraphics{figure_4} Mary swims in still water at 0.85 m s\(^{-1}\). She swims across a straight river which is 60 m wide and flowing at 0.4 m s\(^{-1}\). She sets off from a point \(A\) on the near bank and lands at a point \(B\), which is directly opposite \(A\) on the far bank, as shown in Fig. 4. Find

1. the angle between the near bank and the direction in which Mary swims, [3]
2. the time she takes to cross the river. [3]

\includegraphics{figure_5} A little further downstream a large tree has fallen from the far bank into the river. The river is modelled as flowing at 0.5 m s\(^{-1}\) for a width of 40 m from the near bank, and 0.2 m s\(^{-1}\) for the 20 m beyond this. Nassim swims at 0.85 m s\(^{-1}\) in still water. He swims across the river from a point \(C\) on the near bank. The point \(D\) on the far bank is directly opposite \(C\), as shown in Fig. 5. Nassim swims at the same angle to the near bank as Mary.

1. Find the maximum distance, downstream from \(CD\), of Nassim during the crossing. [5]
2. Show that he will land at the point \(D\). [4]

\includegraphics{figure_1} A girl swims in still water at 1 m s\(^{-1}\). She swims across a river which is 336 m wide and is flowing at 0.6 m s\(^{-1}\). She sets off from a point \(A\) on one bank and lands at a point \(B\), which is directly opposite \(A\), on the other bank as shown in Fig. 1. Find

1. the direction, relative to the earth, in which she swims, [3]
2. the time that she takes to cross the river. [3]

A girl can paddle her canoe at \(5 \text{ m s}^{-1}\) in still water. She wishes to cross a river which is \(100 \text{ m}\) wide and flowing at \(8 \text{ m s}^{-1}\).

1. 1. Write down the angle to the river bank at which the boat must head, in order to cross the river in the least possible time. [1]
   2. Find the acute angle to the river bank at which the boat must head, in order to cross the river by the shortest route. [4]
2. Calculate the times taken for each of the two cases in part (i). [3]