A train at a constant 41.0 km/h moves east for 38 min, then in a direction 44.0° east of due north for 25.0 min, and then west for 59.0 min. What are the (a) magnitude (in km/h) and (b) angle (relative to north, with east of north positive and west of north negative) of its average velocity during this trip?
See my answer to your later post of the same question.
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To find the average velocity of the train, we need to calculate both the magnitude (in km/h) and the angle (relative to north) of its average velocity vector.
Step 1: Calculate the total displacement of the train.
To find the total displacement of the train, we need to calculate the vector sum of its three displacements.
Displacement 1: East for 38 minutes at a constant speed of 41.0 km/h.
Since speed is distance traveled divided by time, the distance traveled in the first displacement is (41.0 km/h) * (38/60) h = 25.7 km. The displacement vector for this leg of the trip is 25.7 km to the east.
Displacement 2: In a direction 44.0° east of due north for 25 minutes.
To find the distance traveled in this displacement, we need to calculate the horizontal and vertical components of the displacement. The horizontal component is (41.0 km/h) * (25/60) h = 17.1 km, and the vertical component is (41.0 km/h) * (25/60) h * sin(44°) ≈ 11.8 km.
The displacement vector can be represented as a horizontal component of +17.1 km and a vertical component of +11.8 km.
Displacement 3: West for 59 minutes at a constant speed of 41.0 km/h.
The distance traveled in this displacement is (41.0 km/h) * (59/60) h = 40.2 km. The displacement vector for this leg of the trip is 40.2 km to the west.
Step 2: Calculate the average velocity vector.
To find the average velocity vector, we need to divide the total displacement of the train by the total time taken.
The total displacement vector is the sum of the three displacements calculated above, which is:
(25.7 km east) + (17.1 km east, 11.8 km north) + (40.2 km west)
The total time taken is (38/60) h + (25/60) h + (59/60) h = 1.03 h.
Dividing the displacement vector by the time taken gives us the average velocity vector.
Step 3: Calculate the magnitude and angle of the average velocity.
(a) The magnitude of the average velocity is the length of the average velocity vector, which can be calculated using the Pythagorean theorem. The horizontal component is (+25.7 km + 17.1 km - 40.2 km) km = +2.6 km and the vertical component is (+11.8 km) km = +11.8 km. Therefore, the magnitude of the average velocity is √((2.6 km)² + (11.8 km)²) ≈ 12.1 km/h.
(b) The angle of the average velocity can be found using trigonometry. The angle (relative to north, with east of north positive and west of north negative) is the arctan of the vertical component divided by the horizontal component. Therefore, the angle is arctan((11.8 km) / (2.6 km)) ≈ 77.4° east of north.
So, the answers to the questions are:
(a) The magnitude of the average velocity is approximately 12.1 km/h.
(b) The angle (relative to north, with east of north positive and west of north negative) of the average velocity is approximately 77.4° east of north.