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: [