A man wishes to cross a river 50m wide which is flowing at 0.5m/s. Given that he can swim at 0.8m/s, find:

Question

(a) The time he takes if he decides to swim straight across the river
(b) The shortest possible time in which he can reach the opposite bank.
Include drawing

Answer 1

To analyze the problem of a man swimming across a river while affected by a current, we will address both parts of the question.

Given:

• Width of the river (W) = 50 m
• Speed of the river (v_r) = 0.5 m/s
• Swimming speed of the man (v_s) = 0.8 m/s

(a) Time taken if he swims straight across the river

When swimming straight across, the swimmer will be affected by the current. His effective velocity will have two components: the component directly across the river and the component downstream due to the current.

1. Time to Cross the River: The time to cross the river directly, without considering the drift due to the current, can be calculated using the formula: \[ \text{Time} = \frac{\text{Width of river}}{\text{Swimming speed across the river}} \] Here, he swims directly across, so we use \(v_s\): \[ \text{Time} = \frac{50 m}{0.8 m/s} = 62.5 s \]

2. Downstream Drift: Meanwhile, during this time, due to the river current, he will drift downstream: \[ \text{Distance drifted downstream} = v_r \times \text{Time} \] \[ \text{Distance drifted downstream} = 0.5 m/s \times 62.5 s = 31.25 m \]

So, if the man swims directly across, he will take 62.5 seconds, and he will also drift 31.25 meters downstream upon reaching the opposite bank.

(b) Shortest possible time to reach the opposite bank

To minimize the time taken across the river considering the current, the swimmer should aim at an angle upstream. This can be analyzed via the concept of relative velocity and vector components.

1. Effective Swimming Direction: To find the optimal angle, we can use the Pythagorean theorem relating his swimming velocity, the current, and the across-river component.

   Let \( v_{y} \) be the effective component of his swimming velocity directed across the river: \[ v_{y} = v_s \cos(\theta) \] and let \( v_{x} \) be the component downstream that counteracts the current: \[ v_{x} = v_s \sin(\theta) \] We require \( v_{x} = v_r \) to counteract the river. Thus: \[ v_s \sin(\theta) = 0.5 m/s \] Substituting the value of \(v_s\): \[ 0.8 \sin(\theta) = 0.5 \] \[ \sin(\theta) = \frac{0.5}{0.8} = 0.625 \] Therefore, calculating \( \theta \): \[ \theta = \arcsin(0.625) \approx 38.68^\circ \]

2. Time to Cross Using Optimal Direction: Now, the effective component of his swimming that helps him cross the river can be derived: \[ v_{y} = 0.8 \cos(\theta) \] Calculating \( \cos(\theta) \): \[ \cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - 0.625^2} \approx \sqrt{1 - 0.390625} \approx \sqrt{0.609375} \approx 0.781 \] Then: \[ v_{y} = 0.8 \times 0.781 \approx 0.625 m/s \]

The time to cross the river is then: \[ \text{Time} = \frac{W}{v_y} = \frac{50 m}{0.625 m/s} = 80 s \]

Thus, the shortest possible time to reach the opposite bank considering the current is 80 seconds.

Summary

(a) The time he takes if he swims straight across the river is 62.5 seconds with a drift of 31.25 meters downstream.
(b) The shortest possible time in which he can reach the opposite bank is 80 seconds.

Drawing (not to scale)

        N
         |
         | \
         |  \
         |   \
         |    \       (Man swims at angle θ)
         |     \
         |      \
         |-------O  ← Current (v_r = 0.5 m/s)
         |     
                   
--------(50m)--------------------S  (Bank opposite to starting point)

• N = North bank (starting point)
• S = South bank (destination)
• O = Starting point
• The angle θ represents the direction upstream he should swim.