A train moving at a constant speed of 61.0 km/h moves east for 35.0 min, then in a direction 50.0° east of due north for 10.0 min, and then west for 49.0 min. What is the average velocity of the train during this run?

What is the magnitude of that velocity?
What is the direction (counterclockwise from east)?

d1 = 61km/h[0o] * (35/60)h = 35.6km[0o].

d2=61km/h[40o] * (10/60)h=10.2km[40o].

d3=61km/h[180o] * (49/60)h=49.8km[180o].

T = (35+10+49)min * 1h/60min = 1.57 h.

X=35.6+10.2*cos40+49.8*cos180=-6.39km.
Y = 10.2*sin40 + 49.8*sin180 = 6.56 km.

tan Ar = Y/X = 6.56/-6.39 = -1.02605
Ar = -45.7o = Reference angle.
A = -45.7 + 180=134.3o, CCW.=Direction

d = Y/sin A = 6.56/sin134.3 = 9.17 km.
V = d/t = 9.17/1.57 = 5.84 km/h.

Why did the train decide to take a detour? Because it wanted to explore its artistic side and try some abstract directions!

To calculate the average velocity of the train, we'll need to break down its movements:

1. The train moves east for 35.0 minutes at a speed of 61.0 km/h.
2. Then it moves 50.0° east of due north for 10.0 minutes.
3. Finally, it moves west for 49.0 minutes.

To find the average velocity, we'll calculate the displacement in both x and y directions.

In the east-west direction:
Time = 35.0 + 49.0 = 84.0 minutes = 1.4 hours
Displacement = speed × time = 61.0 km/h × 1.4 h = 85.4 km west

In the north-south direction:
Time = 10.0 minutes = 0.1667 hours
Displacement = speed × time = 61.0 km/h × 0.1667 h = 10.0 km north

Now we can find the magnitude of the average velocity:
Magnitude = √[(displacement in x direction)² + (displacement in y direction)²]
Magnitude = √[(85.4 km)² + (10.0 km)²]
Magnitude = √(7293.16 km² + 100 km²)
Magnitude ≈ √7393.16 km²
Magnitude ≈ 86.0 km/h

And lastly, to find the direction:
Direction = arctan[(displacement in y direction)/(displacement in x direction)]
Direction = arctan(10.0 km/85.4 km)
Direction ≈ 6.7° counterclockwise from east

So, the average velocity of the train is approximately 86.0 km/h with a direction of about 6.7° counterclockwise from east. I hope this answer gets your train of thought back on track!

To find the average velocity of the train during the entire run, we need to calculate the total displacement and divide it by the total time.

Step 1: Convert time to hours
The time spent moving east is 35.0 minutes, which is equal to 35.0/60 = 0.5833 hours.
The time spent moving in the direction 50.0° east of due north is 10.0 minutes, which is equal to 10.0/60 = 0.1667 hours.
The time spent moving west is 49.0 minutes, which is equal to 49.0/60 = 0.8167 hours.

Step 2: Calculate the displacement in the x-direction (east/west)
Since the train moves east for 35.0 minutes and then moves west for 49.0 minutes, the displacement in the x-direction is 0 km as the eastward and westward displacements cancel each other out.

Step 3: Calculate the displacement in the y-direction (north)

Displacement in the y-direction = distance * sin(angle from north)

The distance traveled in the direction 50.0° east of due north can be calculated by:
distance = speed * time

For the 10 minutes spent traveling in the direction 50.0° east of due north:
distance = 61.0 km/hr * 0.1667 hr
distance = 10.18 km

The displacement in the y-direction is then:
displacement = distance * sin(angle from north)
displacement = 10.18 km * sin(50.0°)
displacement ≈ 7.83 km

So, the displacement in the y-direction is approximately 7.83 km.

Step 4: Calculate the total displacement and time

The total displacement is the vector sum of the x and y displacements. Since the x-direction displacement is 0 km, the total displacement is equal to the displacement in the y-direction.

The total time is the sum of the individual times spent in each direction:
Total time = 0.5833 hr + 0.1667 hr + 0.8167 hr
Total time ≈ 1.5667 hr

Step 5: Calculate average velocity

Average velocity = total displacement / total time
Average velocity = 7.83 km / 1.5667 hr
Average velocity ≈ 5.00 km/h

So, the average velocity of the train during this run is approximately 5.00 km/h.

To find the magnitude of the velocity:
The magnitude of the velocity is the absolute value of the average velocity.
Magnitude of velocity = |5.00 km/h|
Magnitude of velocity = 5.00 km/h

So, the magnitude of the velocity is 5.00 km/h.

To find the direction (counterclockwise from east):
The direction can be found using the inverse tangent function, arctan.
Direction (counterclockwise from east) = arctan(displacement in the y-direction / displacement in the x-direction)

In our case, the displacement in the x-direction is 0 km and the displacement in the y-direction is 7.83 km.
Direction (counterclockwise from east) = arctan(7.83 km / 0 km)
Since the displacement in the x-direction is 0 km, this implies the train has not moved east or west, therefore, it does not have a direction counterclockwise from east.

So, the direction (counterclockwise from east) is undefined.

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

1. First, let's calculate the displacement in the east direction:
The train moves east for 35.0 minutes at a speed of 61.0 km/h.
Displacement = Speed × Time = 61.0 km/h × (35.0 min ÷ 60 min/h) = 35.92 km east

2. Next, let's calculate the displacement in the north direction:
The train moves in a direction 50.0° east of due north for 10.0 minutes.
Displacement = Speed × Time = 61.0 km/h × (10.0 min ÷ 60 min/h) = 10.17 km north

3. Then, let's calculate the displacement in the west direction:
The train moves west for 49.0 minutes at a speed of 61.0 km/h.
Displacement = Speed × Time = 61.0 km/h × (49.0 min ÷ 60 min/h) = 50.58 km west

Now, we can find the total displacement by adding the east and west displacements and subtracting the north displacement:
Total Displacement = (35.92 km east + 10.17 km north) - 50.58 km west
Total Displacement = -4.49 km west

4. Finally, to find the average velocity, we divide the total displacement by the total time taken:
Average Velocity = Total Displacement ÷ Total Time
Total Time = 35.0 min + 10.0 min + 49.0 min = 94.0 min

Average Velocity = -4.49 km ÷ (94.0 min ÷ 60 min/h) = -2.87 km/h

So, the average velocity of the train during this run is -2.87 km/h (west).

Now let's calculate the magnitude of that velocity:
Magnitude of velocity is the absolute value of the average velocity.

Magnitude of velocity = |Average Velocity| = |-2.87 km/h| = 2.87 km/h

Therefore, the magnitude of the velocity of the train during this run is 2.87 km/h.

To find the direction measured counterclockwise from east, we can use trigonometry:

Direction = arctan(north displacement / east displacement)
Direction = arctan(10.17 km / 35.92 km) = 15.92°

Therefore, the direction of the velocity of the train during this run, measured counterclockwise from east, is approximately 15.92°.