A train moving at a constant speed of 60.0 km/h moves east for 40.0 min. then in a direction 50.0 degrees east of north for 20.0 min, and finally west for 50.0 min.
What is the average velocity of the train during this trip?
2006-09-25 07:51:51 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
The key concept here is the difference between speed and velocity. Velocity is a vector, while speed is not. In simple terms, if I run 5 feet to the right, and then 5 feet back to the left, it makes no difference how fast I ran, because I'm back where I started, so my average velocity is zero!
In order for you to find the average velocity, you'll need to work out how far the train is from its starting point, and how long it took to get there -- and then divide one by the other.
1. First the train travels due east for 40.0 min at 60.0 km/h, putting it 40.0 km east.
2. It's *last* step is to move west for 50.0 min, which gives us an overall movement of 10.0 km west.
3. In the middle step, it moves 50.0 degrees east of north for 20.0 min, which requires trigonometry to work out.
Hope that gets you started!
2006-09-25 07:58:15 · answer #1 · answered by Jay H 5 · 1⤊ 0⤋
Since the train is moving at constant speed the distance per unit time traveled in the different directions is the same.The distance traveld in each direction is different.So the average velocity is the speed of the direction of the resultant vector. which is 12.8 km in 12.8 minutes at an angle of 38.9 degrees west or at a bearing of 141.06 degree. The average speed remains the same the velocity has changed.
2006-09-25 08:38:55 · answer #2 · answered by goring 6 · 0⤊ 0⤋
Remember velocity is a vector, both magnitude and direction are important. Find displacement, how far you have traveled from your starting position in a straight line, not necessarily how far on the track you traveled and divide by the total time to find average velocity.
2006-09-25 08:00:26 · answer #3 · answered by msi_cord 7 · 0⤊ 0⤋
average velocity=total dist/total time
total distance=60(2/3)+60tan40*(1/3)+60(50/60)
total time=40+50+20=110/60 hrs
now do the calculation to get the answer
2006-09-25 07:58:54 · answer #4 · answered by raj 7 · 0⤊ 0⤋
t2-t1=40+20+50 min = 1.8333333 hr
y2-y1=20cos50
x2-x1=40+20sin50-50=20(sin50-0.5)
(r2-r1)^2=400(1.25-sin50)
vbar=20*60sqrt(1.25-sin50)/110 @arctan(cos50/(sin50-0.5))
vbar=7.589 km/hr @ 22.484 deg East of North
2006-09-25 08:20:39 · answer #5 · answered by Helmut 7 · 0⤊ 0⤋
60mph @22.49 degrees east of north
2006-09-25 07:58:37 · answer #6 · answered by bruinfan 7 · 0⤊ 0⤋
60.0 km/h d'uh!
2006-09-25 07:58:12 · answer #7 · answered by Andy FF1,2,CrTr,4,5,6,7,8,9,10 5 · 0⤊ 1⤋