A train at a constant 44.0 km/h moves east for 42 min, then in a direction 45.0° east of due north for 17.0 min, and then west for 51.0 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?
d1 = 44km/h * (42/60)h = 30.80 km @ 0o.
d2 = 44km/h * (17/60)h = 12.47 km @ 45o.
d3 = 44km/h * (51/60)h = 37.4 km @ 180o.
a.X=Hor.=30.80+12.47*cos45+37.40*cos180 =2.22 km.
Y = 12.47*sin45+0 = 8.818 km.
D^2 = X^2 + Y^2 = 4.928 + 77.75 = 82.68
D = 9.1 km. = Total distance.
T = 42/60 + 17/60 + 51/60 = 1.833 h.
V = D/T = 9.1km / 1.833h = 4.965 km/h.
b. tanA = Y/X = 8.818/2.22 = 3.97207
A = 75.9o,CCW = 14.13o East of North.
= Direction.
To find the magnitude and angle of the average velocity of the train, we need to calculate the total displacement and the total time taken during the trip.
First, let's break down the train's motion into its components. The train first moves east, then northeast, and finally westward.
1. Eastward motion:
The train travels at a constant speed of 44.0 km/h for 42 minutes. We can convert this time to hours by dividing by 60. So the eastward displacement can be calculated using the formula:
displacement (east) = (velocity) x (time)
displacement (east) = 44.0 km/h x (42/60) h
2. Northeastward motion:
The train moves in a direction 45.0° east of due north for 17.0 minutes. We need to find the northward and eastward components of this motion. The northward component is given by:
displacement (north) = (velocity) x (time) x cos(theta)
displacement (north) = 44.0 km/h x (17/60) h x cos(45°)
Similarly, the eastward component can be calculated using:
displacement (east) = (velocity) x (time) x sin(theta)
displacement (east) = 44.0 km/h x (17/60) h x sin(45°)
3. Westward motion:
Finally, the train moves westward for 51 minutes. To calculate the displacement (west), we use:
displacement (west) = (velocity) x (time)
displacement (west) = 44.0 km/h x (51/60) h
Now, we can find the total displacement by adding the individual displacements of each segment:
total displacement (north) = displacement (north) - displacement (east)
total displacement (north) = [44.0 km/h x (17/60) h x cos(45°)] - [44.0 km/h x (17/60) h x sin(45°)]
total displacement (east) = displacement (east) + displacement (west)
total displacement (east) = [44.0 km/h x (17/60) h x sin(45°)] + [44.0 km/h x (51/60) h]
The total time taken is the sum of the times for each segment:
total time = 42 minutes + 17 minutes + 51 minutes
Finally, we can calculate the magnitude (speed) and angle (direction) of the average velocity using the formula:
average velocity = total displacement / total time
(a) Magnitude of average velocity:
magnitude = square root of [(total displacement (east))^2 + (total displacement (north))^2] / total time
(b) Angle of average velocity:
angle = arctan(total displacement (east) / total displacement (north))
Now, plug in the values and calculate the magnitude and angle of the average velocity of the train during this trip.