Suppose you’re on a boat on a lake with very still water. You walk on the boat at speed of 5 km/hr. The boat is moving at 12 km/hr relative to the water, in a direction that is 135o counterclockwise from the direction you are walking. What is your absolute speed relative to the water?

To find your absolute speed relative to the water, we need to apply vector addition, considering both your speed on the boat and the boat’s speed relative to the water. Let’s break it down step by step.

### Step 1: Represent velocities as vectors
1. **Your speed relative to the boat** is \( 5 \, \text{km/h} \). We can assume you are walking in the positive x-direction, so this vector can be written as:
   \[
   \mathbf{v_{\text{you/boat}}} = 5 \hat{i}
   \]
2. **The boat’s speed relative to the water** is \( 12 \, \text{km/h} \), moving in a direction that is \( 135^\circ \) counterclockwise from your direction of walking. This angle is measured from the positive x-axis (direction of your walking). The components of the boat's velocity vector in terms of x and y can be found using trigonometry:
   \[
   \mathbf{v_{\text{boat/water}}} = 12 \left( \cos(135^\circ) \hat{i} + \sin(135^\circ) \hat{j} \right)
   \]
   Using the values of cosine and sine for \( 135^\circ \):
   \[
   \cos(135^\circ) = -\frac{1}{\sqrt{2}}, \quad \sin(135^\circ) = \frac{1}{\sqrt{2}}
   \]
   So, the boat’s velocity vector becomes:
   \[
   \mathbf{v_{\text{boat/water}}} = 12 \left( -\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} \right)
   \]
   Simplifying:
   \[
   \mathbf{v_{\text{boat/water}}} = -6\sqrt{2} \hat{i} + 6\sqrt{2} \hat{j}
   \]

### Step 2: Add the two velocity vectors
To find your velocity relative to the water, we add your velocity relative to the boat and the boat's velocity relative to the water:
\[
\mathbf{v_{\text{you/water}}} = \mathbf{v_{\text{you/boat}}} + \mathbf{v_{\text{boat/water}}}
\]
\[
\mathbf{v_{\text{you/water}}} = 5 \hat{i} + \left( -6\sqrt{2} \hat{i} + 6\sqrt{2} \hat{j} \right)
\]
Simplifying:
\[
\mathbf{v_{\text{you/water}}} = \left( 5 - 6\sqrt{2} \right) \hat{i} + 6\sqrt{2} \hat{j}
\]

### Step 3: Calculate the magnitude of your velocity
The magnitude of your velocity vector (your absolute speed) is given by:
\[
|\mathbf{v_{\text{you/water}}}| = \sqrt{\left( 5 - 6\sqrt{2} \right)^2 + \left( 6\sqrt{2} \right)^2}
\]
We can now calculate this expression numerically.

Let me compute this for you.Your absolute speed relative to the water is approximately \( 9.17 \, \text{km/h} \).

Would you like more details or have any questions about this solution?

Here are five related questions to expand on this topic:
1. How would the result change if the boat's speed was reduced to 8 km/h?
2. What if you were walking at an angle of 45° relative to the boat instead of straight ahead?
3. How would your speed change if the boat were moving directly opposite to your walking direction?
4. What is the significance of vector addition in real-life problems like this?
5. How does the angle between two moving objects affect their relative velocities?

**Tip:** When dealing with velocities, always break them into components to simplify vector addition.

Suppose you’re on a boat on a lake with very still water. You walk on the boat at speed of 5 km/hr. The boat is moving at 12 km/hr relative to the water, in a direction that is 135° counterclockwise from the direction you are walking. What is your absolute speed relative to the water?

This problem demonstrates how to calculate your absolute speed relative to water using vector addition and trigonometry. You are walking at 5 km/h on a boat moving at 12 km/h at a 135° angle. Learn to apply vector components and the Pythagorean theorem to find the result, suitable for students in Grades 11-12.

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