A swimmer starts swimming north at a velocity of 5 m/s while the current is taking her at 8 m/s at an angle of 300 degrees. How far is she from the start after 5 minutes?

To solve this problem, we need to determine the swimmer's total displacement after 5 minutes by considering both her own velocity and the velocity of the current.

### Step 1: Decompose the current's velocity
The current is moving at 8 m/s at an angle of 300°. To break this down into its horizontal (\(x\)) and vertical (\(y\)) components, we use trigonometry:

- \(V_{x} = 8 \cos(300^\circ)\)
- \(V_{y} = 8 \sin(300^\circ)\)

Since \( \cos(300^\circ) = \frac{1}{2} \) and \( \sin(300^\circ) = -\frac{\sqrt{3}}{2} \), we get:
- \(V_{x} = 8 \times \frac{1}{2} = 4 \, \text{m/s}\)
- \(V_{y} = 8 \times -\frac{\sqrt{3}}{2} = -4\sqrt{3} \approx -6.93 \, \text{m/s}\)

### Step 2: Combine the swimmer's velocity and the current's velocity
The swimmer is moving north (in the positive \(y\)-direction) at 5 m/s, so her total velocity in the \(y\)-direction is the sum of her velocity and the \(y\)-component of the current:

- \(V_{y, \text{total}} = 5 + (-6.93) = -1.93 \, \text{m/s}\)

In the \(x\)-direction, only the current is contributing to the velocity:

- \(V_{x, \text{total}} = 4 \, \text{m/s}\)

### Step 3: Find the total displacement after 5 minutes
The swimmer is moving for 5 minutes, which is 300 seconds. Now, we calculate her displacement in the \(x\) and \(y\) directions:

- \(D_{x} = V_{x, \text{total}} \times t = 4 \times 300 = 1200 \, \text{m}\)
- \(D_{y} = V_{y, \text{total}} \times t = -1.93 \times 300 = -579 \, \text{m}\)

### Step 4: Calculate the total displacement
The total displacement is the magnitude of the resultant vector:

\[
D_{\text{total}} = \sqrt{D_{x}^2 + D_{y}^2} = \sqrt{1200^2 + (-579)^2} = \sqrt{1440000 + 335241} = \sqrt{1775241} \approx 1332.26 \, \text{m}
\]

Thus, the swimmer is approximately **1332.26 meters** from her starting point after 5 minutes.

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Would you like further details or have any questions?

Here are 5 related questions:
1. How would the problem change if the current's velocity was 10 m/s?
2. What if the swimmer were swimming south instead of north?
3. How would you calculate the time taken to reach a certain distance?
4. How would the calculation change if the current's direction was 45 degrees?
5. How does the angle of the current affect the displacement?

**Tip:** Always break down vectors into components when dealing with motion at angles.

Learn how to calculate the displacement of a swimmer moving at 5 m/s while a current of 8 m/s at 300° acts on her. This detailed solution breaks down the vector components using trigonometry and applies the Pythagorean Theorem to find the total displacement of approximately 1332.26 meters after 5 minutes. Suitable for Grades 10-12.

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