A swimmer orients herself perpendicular to the parallel banks of a river. If the swimmer’s velocity is 2 $ % and the velocity of the current is 0.5 $ % , What will her resultant velocity? What is her direction (in degrees)? How far will she actually have to swim to get to the other side if the banks of the river are 50m apart?

To solve this problem, we can break it down into three parts: 

1. **Find the resultant velocity** (magnitude and direction),
2. **Find the angle of the resultant velocity** (direction),
3. **Calculate how far the swimmer will actually swim to cross the river**.

### Step 1: Resultant Velocity
The swimmer's velocity is perpendicular to the current, which means we can treat this as a right triangle problem. The swimmer's velocity $ v_s = 2 \, \text{m/s} $ and the current's velocity $ v_r = 0.5 \, \text{m/s} $. The resultant velocity $ v_{\text{res}} $ is the vector sum of these two velocities. Using the Pythagorean theorem:

\[
v_{\text{res}} = \sqrt{v_s^2 + v_r^2} = \sqrt{2^2 + 0.5^2} = \sqrt{4 + 0.25} = \sqrt{4.25} \approx 2.06 \, \text{m/s}
\]

So, the resultant velocity is approximately $ 2.06 \, \text{m/s} $.

### Step 2: Direction (Angle)
The direction of the resultant velocity can be found using trigonometry. Specifically, we use the tangent function because we are given the opposite (velocity of the current) and adjacent (velocity of the swimmer):

\[
\tan(\theta) = \frac{v_r}{v_s} = \frac{0.5}{2} = 0.25
\]

Now, solve for $ \theta $ (the angle relative to the bank):

\[
\theta = \tan^{-1}(0.25) \approx 14.04^\circ
\]

Thus, the swimmer's direction is approximately $ 14.04^\circ $ downstream from her intended path.

### Step 3: Distance Actually Swum
The swimmer's path across the river is not straight across because the current pushes her downstream. The actual distance she swims can be calculated by using the width of the river as the opposite side of a right triangle and her resultant velocity direction.

The width of the river is 50 meters, so the distance she will actually swim is the hypotenuse of the triangle:

\[
\text{Distance} = \frac{50}{\cos(\theta)} = \frac{50}{\cos(14.04^\circ)} \approx \frac{50}{0.9703} \approx 51.53 \, \text{m}
\]

Thus, the swimmer will actually have to swim approximately $ 51.53 \, \text{m} $.

### Summary:
1. **Resultant velocity**: $ 2.06 \, \text{m/s} $
2. **Direction**: $ 14.04^\circ $ downstream
3. **Actual distance swum**: $ 51.53 \, \text{m} $

Would you like more details on any step or have further questions?

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Here are five related questions:
1. How would the resultant velocity change if the current’s speed doubled?
2. If the river's width doubled, how would the actual distance swum change?
3. What if the swimmer's velocity was reduced to 1.5 m/s instead of 2 m/s?
4. How much time will the swimmer take to cross the river with these given velocities?
5. If the current's speed was 1 m/s, what would the new angle of the resultant velocity be?

**Tip:** When dealing with vector problems, always break them down into components (horizontal and vertical) for easier calculations.

A swimmer orients herself perpendicular to the parallel banks of a river. If the swimmer’s velocity is 2 m/s and the velocity of the current is 0.5 m/s, what will her resultant velocity be? What is her direction (in degrees)? How far will she actually have to swim to get to the other side if the banks of the river are 50m apart?

Learn how to calculate the resultant velocity and direction of a swimmer in a river with a current of 0.5 m/s and swimmer velocity of 2 m/s. Solve this using the Pythagorean theorem and trigonometry to find the actual distance swum when the river is 50 meters wide. Suitable for Grades 10-12.

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