II) (a) At what upstream angle must the swimmer in Problem 67 aim, if she is to arrive at a point
directly across the stream? (b) How long will it take her? [GSE 3-68] the swimmer moves at .6 m/s the river flows at .5 m/s and the river is 55m wide

Let's break down the problem. The swimmer needs to reach a point directly across the stream, so her swimming direction must counteract the current of the river.

### Part (a): Finding the Upstream Angle

The swimmer's velocity is \( 0.6 \, \text{m/s} \) relative to still water, and the river's velocity is \( 0.5 \, \text{m/s} \).

To arrive directly across, the horizontal (downstream) component of her swimming velocity must cancel the velocity of the river. If she swims at an angle \( \theta \) upstream, the horizontal component of her velocity, \( v_x \), will be:
\[
v_x = v_s \sin(\theta) = 0.6 \sin(\theta)
\]
where \( v_s = 0.6 \, \text{m/s} \) is the swimmer's speed. This horizontal component must equal the velocity of the river, \( v_r = 0.5 \, \text{m/s} \), so:
\[
0.6 \sin(\theta) = 0.5
\]
Solving for \( \sin(\theta) \):
\[
\sin(\theta) = \frac{0.5}{0.6} = 0.8333
\]
Thus, \( \theta \) is:
\[
\theta = \arcsin(0.8333) \approx 56.44^\circ
\]
So, the swimmer must aim approximately **56.44° upstream**.

### Part (b): Time to Cross the River

The width of the river is \( 55 \, \text{m} \), and the swimmer’s velocity perpendicular to the river’s current (the vertical component) is:
\[
v_y = v_s \cos(\theta) = 0.6 \cos(56.44^\circ)
\]
First, let's calculate \( \cos(56.44^\circ) \):
\[
\cos(56.44^\circ) \approx 0.5556
\]
Thus, the vertical component of her velocity is:
\[
v_y = 0.6 \times 0.5556 \approx 0.3334 \, \text{m/s}
\]
Now, the time \( t \) to cross the river is given by:
\[
t = \frac{\text{distance}}{\text{speed}} = \frac{55}{0.3334} \approx 165 \, \text{seconds}
\]
Thus, it will take her approximately **165 seconds** (or 2 minutes and 45 seconds) to cross the river.

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Would you like more details or clarification on any part of the solution?

### Related Questions:
1. How would the angle change if the swimmer’s speed were increased to 1 m/s?
2. What if the river's flow increased to 0.8 m/s? How would that affect the angle?
3. How much time would the swimmer save if she aimed directly downstream?
4. What is the swimmer’s resultant velocity when swimming at this upstream angle?
5. How would the problem change if the river were 100 meters wide?

#### Tip:
Whenever you're solving for angles involving velocities and currents, break the velocity into horizontal and vertical components to simplify your calculations.

II) (a) At what upstream angle must the swimmer in Problem 67 aim, if she is to arrive at a point directly across the stream? (b) How long will it take her? The swimmer moves at 0.6 m/s, the river flows at 0.5 m/s, and the river is 55 m wide.

Find out how to calculate the upstream angle and time needed for a swimmer to cross a 55-meter wide river with a 0.6 m/s swimming speed and a 0.5 m/s current. This problem involves trigonometric principles and vector decomposition. Suitable for students in Grades 10-12.

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