To solve this problem, we need to analyze the situation using vector components of the man's swimming speed and the river current.
Given:
- Width of the river (W) = 50 m
- Speed of the river (Vr) = 0.5 m/s
- Speed of the man swimming (Vm) = 0.5 m/s
(a) Time taken to swim straight across the river
When the man swims directly across the river, he will have a resultant motion that is a combination of his swimming speed and the river's current. Since he swims straight across, we only consider the width of the river for the time calculation and ignore the downstream movement caused by the current.
- Time = Distance / Speed
The distance he needs to swim directly across the river is the width of the river (50 m).
\[ \text{Time}_\text{straight} = \frac{W}{V_m} = \frac{50 \text{ m}}{0.5 \text{ m/s}} = 100 \text{ s} \]
Time taken if he swims straight across the river = 100 seconds.
(b) The shortest possible time to reach the opposite bank
To find the shortest time to cross the river, the man should aim upstream at an angle so that the downstream drift due to the current is minimized. He needs to swim at an angle \(\theta\) to compensate for the current.
Using trigonometry, we can set up the following relationships:
- The component of the man's swimming speed that is toward the river's width (perpendicular to the current) is \(V_m \cdot \sin(\theta)\).
- The component against the current (parallel to the bank) should equal the current speed, i.e., \(V_m \cdot \cos(\theta) = V_r\).
From the second equation:
\[ 0.5 \cdot \cos(\theta) = 0.5 \] \[ \cos(\theta) = 1 \implies \theta = 0 \text{ (which is swimming straight across)} \]
However, to minimize the total time:
Let’s calculate the angle:
- The man's effective speed across the width of the river must remain effective while countering the current.
We know: \[ V_m \cdot \cos(\theta) = V_r \] Thus: \[ \cos(\theta) = \frac{0.5}{0.5} = 1 \implies \theta = 0 \]
This implies that if \(V_m\) is equal to \(V_r\), he can swim at an angle of \(45^\circ\) to most efficiently use both velocities \(V_m\) towards the width and \(V_r\) along.
- The width is still \(50 m\) and we know swimming speed effectively will not change.
Finally, to minimize the time taken, he would need to swim at an angle where these components adjust to hitting directly at the perpendicular angle to cross:
Using the Pythagorean theorem: \[ V_t = \sqrt{V_m^2 - V_r^2} = \sqrt{(0.5)^2 - (0.5)^2} = 0.5 m/s \]
Thus, swimming speed remains \( \sqrt{(0.5)^2 + (0.5)^2} = 0.707 \text{ m/s}\) minimizing to \(45^\circ\).
- Time = Distance / Effective Speed toward opposite bank:
\[ \text{Effective Speed} = V_m \cdot \sin(45^{\circ}) = 0.5 \cdot \frac{\sqrt{2}}{2} = 0.353 \text{ m/s} \]
Thus, the time to cross:
\[ \text{Time} = \frac{50}{0.353} \approx 142 \text{ s} \]
Summary:
- (a) Time taken to swim straight across = 100 seconds.
- (b) Shortest possible time = Approx. 142 seconds.
Drawing
A visual depiction can be made with the river as a rectangle, showing the swimmer at angles opposing the current and using the coordinate planes to depict velocity segments and resultant paths.
- Draw the river (50 m wide horizontally).
- Indicate the swimmer's swimming direction at an angle against the current to represent compensating versus directly towards the bank.
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