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 °
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?
1 answer
1 answer