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Let's go through these problems step by step.

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**Problem 5:**

A helicopter flies west at 50 km/h, and a 10 km/h wind is blowing south. We need to determine the resultant velocity vector (magnitude and direction).

### Solution:

1. The helicopter's velocity:
   - Westward velocity (\( v_w \)) = 50 km/h (in the negative x-direction)
   
2. Wind velocity:
   - Southward velocity (\( v_s \)) = 10 km/h (in the negative y-direction)

This forms a right-angled triangle, where the components are:
- Horizontal component: \( 50 \, \text{km/h} \)
- Vertical component: \( 10 \, \text{km/h} \)

#### Magnitude of the resultant velocity:
Using the Pythagorean theorem:
\[
v = \sqrt{v_w^2 + v_s^2} = \sqrt{50^2 + 10^2} = \sqrt{2500 + 100} = \sqrt{2600} \approx 50.99 \, \text{km/h}
\]

#### Direction of the resultant velocity:
The angle (\( \theta \)) south of west can be found using:
\[
\theta = \tan^{-1}\left(\frac{v_s}{v_w}\right) = \tan^{-1}\left(\frac{10}{50}\right) = \tan^{-1}(0.2) \approx 11.31^\circ
\]

So, the helicopter is traveling at **50.99 km/h** at an angle of **11.31° south of west**.

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**Problem 6:**

Mr. Gordon runs:
- 20 m North
- 2 m East
- 5 m South
- 12 m West

We need to find the resultant displacement vector.

### Solution:

1. **North-South direction (y-axis):**
   - \( 20 \, \text{m north} - 5 \, \text{m south} = 15 \, \text{m north} \)

2. **East-West direction (x-axis):**
   - \( 2 \, \text{m east} - 12 \, \text{m west} = 10 \, \text{m west} \)

Thus, Mr. Gordon’s displacement components are:
- \( x = -10 \, \text{m} \) (west)
- \( y = 15 \, \text{m} \) (north)

#### Magnitude of the resultant displacement:
Using the Pythagorean theorem:
\[
d = \sqrt{(-10)^2 + 15^2} = \sqrt{100 + 225} = \sqrt{325} \approx 18.03 \, \text{m}
\]

#### Direction of the displacement:
The angle \( \theta \) west of north can be found using:
\[
\theta = \tan^{-1}\left(\frac{10}{15}\right) = \tan^{-1}(0.6667) \approx 33.69^\circ
\]

So, Mr. Gordon’s displacement is approximately **18.03 m at an angle of 33.69° west of north**.

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**Problem 7:**

Usain Bolt runs:
- 590 m west
- 43000 cm east (which is 430 m)
- 1.2 km west (which is 1200 m)

We need to find his resultant displacement.

### Solution:

1. Convert all units to meters:
   - \( 590 \, \text{m west}, 430 \, \text{m east}, 1200 \, \text{m west} \)

2. **Net displacement in the east-west direction (x-axis):**
   - \( 590 \, \text{m west} + 1200 \, \text{m west} - 430 \, \text{m east} = 1360 \, \text{m west} \)

Thus, Usain Bolt’s displacement is **1360 m west**.

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Do you want more details on any of the steps? Here are 5 related questions:

1. How would the displacement change in Problem 5 if the wind was blowing east instead of south?
2. How would you calculate the velocity of the helicopter in Problem 5 if it were traveling north instead of west?
3. In Problem 6, what would Mr. Gordon's displacement be if he ran an additional 10 meters east?
4. How would you approach Problem 6 if there were diagonal movements, like northeast?
5. How does changing the order of directions in Problem 7 affect the total displacement?

**Tip**: When working with vectors, always keep track of both magnitude and direction!

5. A helicopter directs its flight west at 50 km/hr. A 10 km/hr wind to the south angles its flight. At what velocity (magnitude and direction) is the helicopter traveling? 6. Mr. Gordon is trying to catch his dog, Merlin, who escaped outside. He runs 20 m North, then 2 m East, then 5 m South, and then 12 m West. What is the resultant vector of Mr. Gordon's displacement? 7. Usain Bolt runs 590 m west, 43000 cm east, and 1.2 km west. Calculate his displacement (magnitude and direction) in meters.

Learn how to calculate the resultant velocity of a helicopter experiencing wind, Mr. Gordon's displacement vector after chasing his dog, and Usain Bolt's displacement after running in different directions. Problems involve vectors, displacement, velocity, and trigonometry, suitable for Grades 9-12.

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