A boat heads north at 4 m/s for one hour, then turns and goes southeast (at 450 to the north) at 3 m/s for 45 minutes. a) How far has the boat gone? b) How far is it from its starting location?

Let's solve this problem step-by-step.

### a) Total Distance Traveled by the Boat

1. **Distance traveled while heading north:**

   - Speed: \(4 \, \text{m/s}\)
   - Time: 1 hour = \(3600 \, \text{s}\)
   
   Distance, \(d_1 = \text{speed} \times \text{time} = 4 \, \text{m/s} \times 3600 \, \text{s} = 14400 \, \text{m}\).

2. **Distance traveled while heading southeast (at \(45^\circ\) to the north):**

   - Speed: \(3 \, \text{m/s}\)
   - Time: 45 minutes = \(2700 \, \text{s}\)
   
   Distance, \(d_2 = \text{speed} \times \text{time} = 3 \, \text{m/s} \times 2700 \, \text{s} = 8100 \, \text{m}\).

3. **Total distance traveled:**

   \[
   d_{\text{total}} = d_1 + d_2 = 14400 \, \text{m} + 8100 \, \text{m} = 22500 \, \text{m}.
   \]

### b) Displacement from the Starting Location

To find the boat's displacement, we consider the vector components of its journey:

1. **Northward journey:**

   The boat first travels directly north, so the displacement vector after this part is \(14400 \, \text{m}\) in the north direction.

2. **Southeast journey:**

   For the second part, where the boat goes southeast at a \(45^\circ\) angle to the north:

   - The displacement components are calculated using trigonometry:
     \[
     \text{East component} = d_2 \cdot \cos(45^\circ) = 8100 \cdot \frac{\sqrt{2}}{2} \approx 5720.57 \, \text{m},
     \]
     \[
     \text{South component} = d_2 \cdot \sin(45^\circ) = 8100 \cdot \frac{\sqrt{2}}{2} \approx 5720.57 \, \text{m}.
     \]

   - Since the boat is traveling southeast, the east component is positive, and the south component is in the opposite direction of the north, making it negative relative to the first part of the journey.

3. **Net displacement components:**

   - North-South component (N): \(14400 \, \text{m} - 5720.57 \, \text{m} = 8679.43 \, \text{m}\)
   - East-West component (E): \(5720.57 \, \text{m}\)

4. **Magnitude of the resultant displacement:**

   \[
   \text{Displacement} = \sqrt{(8679.43 \, \text{m})^2 + (5720.57 \, \text{m})^2}.
   \]

   Calculating this:

\[
\text{Displacement} = \sqrt{8679.43^2 + 5720.57^2} \, \text{m}.
\]

Let's compute this step-by-step.The boat's displacement from its starting location is approximately **10,395.07 meters**.

### Summary of Results:
a) The total distance traveled by the boat is **22,500 meters**.

b) The distance from its starting location (displacement) is approximately **10,395.07 meters**.

Would you like more details or have any questions about these calculations?

### Related Questions:
1. How would the displacement change if the boat traveled east instead of southeast?
2. What would be the effect on the total distance if the boat traveled at a different speed?
3. How would you calculate the average speed of the boat for the entire journey?
4. What if the boat made a third turn; how would that affect the displacement?
5. How can we represent this problem graphically using vector diagrams?

**Tip:** When dealing with displacement, always consider the direction of each vector component to correctly compute the resultant vector.

A boat heads north at 4 m/s for one hour, then turns and goes southeast (at 45° to the north) at 3 m/s for 45 minutes. a) How far has the boat gone? b) How far is it from its starting location?

This problem involves calculating the total distance a boat travels and its displacement using vector addition and trigonometric principles. Explore key concepts like kinematics, the Pythagorean theorem, and trigonometric ratios in this step-by-step solution. Ideal for students in Grades 10-12.

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