You enter an antique classic car road rally with your 1956 Studebaker Golden Hawk. The rally course consists of the following segments: travel north at 28.0 m/s for 30.0 min, then east at 36.0 m/s for 40.0 min, and finally northeast at 39.0 m/s for 50.0 min. For the entire trip, determine the following.
(a) displacement vector (magnitude and direction) and distance traveled
displacement vector magnitude    	  m
displacement vector direction	  ° North of East
distance	  m

(b) average velocity vector (magnitude and direction) and average speed
average velocity vector magnitude    	  m/s
average velocity vector direction	  ° North of East
average speed	  m/s

Let's break down the problem into parts to determine the displacement vector, distance traveled, and the velocity quantities. 

### Step-by-Step Breakdown:

#### (a) Displacement Vector and Distance Traveled

1. **Time and Velocity for Each Segment:**
   - **North Segment:**
     - Speed = \( 28.0 \, \text{m/s} \)
     - Time = \( 30.0 \, \text{min} = 1800 \, \text{seconds} \)
     - Distance traveled = \( 28.0 \, \text{m/s} \times 1800 \, \text{s} = 50400 \, \text{m} \)

   - **East Segment:**
     - Speed = \( 36.0 \, \text{m/s} \)
     - Time = \( 40.0 \, \text{min} = 2400 \, \text{seconds} \)
     - Distance traveled = \( 36.0 \, \text{m/s} \times 2400 \, \text{s} = 86400 \, \text{m} \)

   - **Northeast Segment (45° angle):**
     - Speed = \( 39.0 \, \text{m/s} \)
     - Time = \( 50.0 \, \text{min} = 3000 \, \text{seconds} \)
     - Distance traveled = \( 39.0 \, \text{m/s} \times 3000 \, \text{s} = 117000 \, \text{m} \)

2. **Component Form of the Displacement:**
   - **North Component** of displacement = North segment + North component of the Northeast segment:
     \[
     D_y = 50400 \, \text{m} + (117000 \, \text{m} \times \sin(45^\circ))
     \]
   
   - **East Component** of displacement = East segment + East component of the Northeast segment:
     \[
     D_x = 86400 \, \text{m} + (117000 \, \text{m} \times \cos(45^\circ))
     \]
   
3. **Magnitude of the Displacement Vector:**
   The magnitude of the displacement vector is the hypotenuse of the right triangle formed by the north and east components:
   \[
   \text{Displacement} = \sqrt{D_x^2 + D_y^2}
   \]

4. **Direction of the Displacement Vector:**
   The direction of the displacement vector (angle north of east) is given by:
   \[
   \theta = \tan^{-1}\left(\frac{D_y}{D_x}\right)
   \]

5. **Total Distance Traveled:**
   The total distance traveled is the sum of the distances in each segment:
   \[
   \text{Total distance} = 50400 \, \text{m} + 86400 \, \text{m} + 117000 \, \text{m}
   \]

#### (b) Average Velocity Vector and Average Speed

1. **Average Speed:**
   Average speed is the total distance traveled divided by the total time:
   \[
  

You enter an antique classic car road rally with your 1956 Studebaker Golden Hawk. The rally course consists of the following segments: travel north at 28.0 m/s for 30.0 min, then east at 36.0 m/s for 40.0 min, and finally northeast at 39.0 m/s for 50.0 min. For the entire trip, determine the following: (a) displacement vector (magnitude and direction) and distance traveled, (b) average velocity vector (magnitude and direction) and average speed.

Solve a detailed car rally problem involving displacement and velocity vectors. Travel through north, east, and northeast directions, calculate total distance, displacement magnitude, direction, average velocity, and speed. Perfect for high school students in Grades 10-12, covering kinematics, trigonometry, and vector analysis.

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