A train at a constant 43.0 km/h moves east for 20.0 min, then in a direction 54.0° east of due north for 25.0 min, and then west for 30.0 min. What are the (a) magnitude and (b) angle (relative to east) of its average velocity during this trip?
Average velocity = (change in position)/time
Assume initial position = (0,0)
First determine the displacement of the train at the end of the trip:
movements(N):
1. 43 km/h for 20 min = 43/3 km due east
2. 43 km/h for 25 min at N54E
= 43*25/60 km at CCW 36° from due east
3. 43 km/h for 30 min. due west
= 43/2 km at 180 CCW from due east
Next, sum the E and N components using sine and cosine
Set dist=actual distance
X=component due east=dist*cos(t)
Y=component due north=dist*sin(t)
N Dist X Y
1 14.3 14.3 0
2 17.9 14.5 10.5
3 21.5 -21.5 0
Final position can be obtained by adding the X and Y components, namely
(7.3,10.5)
total time = 20+25+30 = 75 min. = 1.25 h
Change in position
= (7.3,10.5)-(0,0)
= (7.3, 10.5)
Magnitude of average velocity
=sqrt(7.3^2+10.5^2)/1.25
=12.9/1.25 km/h
=10.3 km/h
Direction:
atan2 (10.5,7.3)=55 °
To find the average velocity of the train during the trip, we need to calculate the displacement and the total time taken.
First, let's calculate the displacement in the east direction.
Distance = Speed × Time
= 43.0 km/h × (20.0 min / 60 min)
= 14.33 km
Next, let's calculate the displacement in the north direction.
Distance = Speed × Time
= 43.0 km/h × (25.0 min / 60 min)
= 17.92 km
Since the train is moving in a direction 54.0° east of due north, we need to find the east and north components of this displacement.
East Component = Displacement × cos(angle)
= 17.92 km × cos(54.0°)
= 10.27 km
North Component = Displacement × sin(angle)
= 17.92 km × sin(54.0°)
= 13.41 km
Next, let's calculate the displacement in the west direction.
Distance = Speed × Time
= 43.0 km/h × (30.0 min / 60 min)
= 21.50 km
To find the total displacement, we need to consider the directions. Since the train initially moves east and then moves west, the net displacement will be the difference between the east and west components.
Net East Component = 14.33 km - 21.50 km
= -7.17 km (westward)
Net North Component = 13.41 km (northward)
Now, let's calculate the magnitude of the net displacement.
Magnitude = sqrt((Net East Component)^2 + (Net North Component)^2)
= sqrt((-7.17 km)^2 + (13.41 km)^2)
= sqrt(51.47 km^2 + 179.88 km^2)
= sqrt(231.35 km^2)
= 15.21 km
To find the angle of the average velocity relative to the east, we can use the inverse tangent function.
Angle = atan(Net North Component / Net East Component)
= atan(13.41 km / (-7.17 km))
= atan(-1.868)
= -60.6°
(a) The magnitude of the average velocity during the trip is 15.21 km.
(b) The angle (relative to east) of the average velocity during the trip is -60.6°.
To find the magnitude and angle of the average velocity, we need to calculate the displacement vector for each segment of the trip and then find the resultant vector.
First, let's calculate the displacement vector for each segment of the trip.
1. Eastward Segment (20.0 min):
Since the train is moving east at a constant speed of 43.0 km/h, its displacement is given by:
Displacement = Velocity x Time = (43.0 km/h) x (20.0 min) x (1 h/60 min) = 14.3 km
2. Northeastward Segment (25.0 min):
The train is moving in a direction 54.0° east of due north. To calculate the displacement vector, we need to resolve the velocity into its north and east components.
North Component = Velocity x cos(angle) = 43.0 km/h x cos(54.0°)
East Component = Velocity x sin(angle) = 43.0 km/h x sin(54.0°)
Displacement = Velocity x Time = (North Component) x (25.0 min) x (1 h/60 min) + (East Component) x (25.0 min) x (1 h/60 min)
3. Westward Segment (30.0 min):
Similar to the eastward segment, the displacement is given by:
Displacement = Velocity x Time = (43.0 km/h) x (30.0 min) x (1 h/60 min) = 21.5 km (negative value since it's westward)
Now, to find the magnitude and angle of the average velocity, we need to calculate the resultant displacement vector.
Resultant Displacement = Displacement1 + Displacement2 + Displacement3
Once we have the resultant displacement, we can calculate the magnitude and angle using the formulas:
Magnitude = sqrt(x^2 + y^2)
Angle = arctan(y/x)
Plug in the values and calculate the magnitude and angle.