A train moving at a constant speed of 68.0 km/h moves east for 33.0 min, then in a direction 40.0° east of due north for 30.0 min, and then west for 59.0 min. What is the average velocity of the train during this run?

Magnitude
km/h

The key is finding the final position.

1) It goes 68*33/60 km east
2) it goes 68*30/60 40 deg E of N. Converting that to E and N displacements..
it goes 60(30/60)sin40 East
it goes 60(30/60)cos40 North
3) it goes E for 68(-59/60) Notice the negative sign which changes West to East.

Now add up the East components.
That gives you the net East.
Now add up the N components.

Now use Pythogerean theorum
displacement=sqrt(N^2 +E^2)

Magnitude avg velocity=displacement/timetotal.

To find the average velocity of the train during this run, we need to calculate the total displacement (change in position) of the train and divide it by the total time taken.

First, let's break down the train's motion into its components.

1. Eastward motion: The train moves east for 33.0 minutes. The speed is given as 68.0 km/h, which means the distance covered in this leg is:

Distance = Speed * Time
Distance = 68.0 km/h * (33.0 min / 60 min)
Distance = 37.4 km

Since the motion is purely eastward, the displacement is in the positive x-direction.

2. Northward motion: The train moves in a direction 40.0° east of due north for 30.0 minutes. To calculate the displacement, we need to break it down into its x and y components. The distance traveled in this leg is:

Distance = Speed * Time
Distance = 68.0 km/h * (30.0 min / 60 min)
Distance = 34.0 km

The x-component of the displacement is:
Displacement x = Distance * cos(40°)
Displacement x = 34.0 km * cos(40°)
Displacement x ≈ 25.94 km (positive x-direction)

The y-component of the displacement is:
Displacement y = Distance * sin(40°)
Displacement y = 34.0 km * sin(40°)
Displacement y ≈ 21.90 km (positive y-direction)

3. Westward motion: The train moves west for 59.0 minutes. The distance covered in this leg is:

Distance = Speed * Time
Distance = 68.0 km/h * (59.0 min / 60 min)
Distance = 67.57 km

Since the motion is purely westward, the displacement is in the negative x-direction.

Now, we can calculate the total displacement:

Total Displacement (Δx) = Final x-position - Initial x-position
Total Displacement (Δx) = (-67.57 km) - (25.94 km)
Total Displacement (Δx) = -93.51 km

Total Displacement (Δy) = Final y-position - Initial y-position
Total Displacement (Δy) = (+21.90 km) - 0 km
Total Displacement (Δy) = +21.90 km

Now, we can calculate the magnitude of the total displacement:

Magnitude of Displacement = sqrt(Δx^2 + Δy^2)
Magnitude of Displacement = sqrt((-93.51 km)^2 + (21.90 km)^2)
Magnitude of Displacement ≈ 95.84 km

Finally, we can calculate the average velocity:

Average Velocity = Total Displacement / Total Time

Total Time = 33.0 min + 30.0 min + 59.0 min
Total Time = 122.0 min

Average Velocity = Magnitude of Displacement / Total Time
Average Velocity ≈ 95.84 km / (122.0 min / 60 min/h)
Average Velocity ≈ 47.42 km/h

Therefore, the average velocity of the train during this run is approximately 47.42 km/h.