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?

Vavg. = V1 + V2 + V3,

Vavg.=41km/h @ 0Deg + 41km/h @ 46Deg. +
41km/h @ 180Deg.

X = 41*cos(0) + 41*cos46 + 41*cos180 =
28.5 km/h.
Y = 41*sin(0) + 41*sin46 + 41*sin180 =
29.5 km/h.

tanA = Y/x = 29.5 / 28.5 = 1.03484,
A = 46 Deg.,CCW=44 Deg. East of North.

V = X/cosA = 28.5 / cos46 = 41 km/h.

V = 41 km/h @ 44 Deg. East of North

CORRECTION:

See later post: Sun,2-9-14, 6:16 AM.

To find the average velocity, we need to calculate the total displacement and total time.

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

Component 1: Eastward motion for 38 min at a speed of 41.0 km/h.
This results in a displacement in the x-direction (east) equal to (41.0 km/h) * (38 min) = 1558 km.

Component 2: Motion in a direction 44.0° east of due north for 25.0 min.
To find the displacement in the y-direction (north), we need to calculate the vertical component of the velocity. The vertical component of the velocity can be calculated as (41.0 km/h) * sin(44.0°) = 27.94 km/h.
Now, we can calculate the displacement in the y-direction by multiplying the vertical component by the time: (27.94 km/h) * (25 min) = 698.5 km.

Component 3: Westward motion for 59.0 min at a speed of 41.0 km/h.
This results in a displacement in the x-direction (west) equal to (41.0 km/h) * (59 min) = -2419 km since west is in the negative x-direction.

Now, let's calculate the total displacement by adding both the x and y components together:
Total displacement in the x-direction = 1558 km - 2419 km = -861 km (negative because it is in the west direction)
Total displacement in the y-direction = 698.5 km.

Next, we need to find the total time taken. The train traveled for 38 min + 25 min + 59 min = 122 min.

Now, we can calculate the average velocity.
The magnitude of the average velocity is given by the total displacement divided by the total time:
Magnitude = (sqrt((-861 km)^2 + (698.5 km)^2)) / (122 min) ≈ 9.72 km/h.

To find the angle of the average velocity relative to North, we can use the inverse tangent function:
Angle = atan2(vertical displacement, horizontal displacement) = atan2(698.5 km, -861 km) ≈ -38.1°.

Therefore, the answers are:
(a) The magnitude of the average velocity is approximately 9.72 km/h.
(b) The angle of the average velocity relative to North is approximately -38.1°.