Ok, I'm not here so that people can do my homework; I just need some help so please don't say I just want answers. I'm having trouble with this question on Relative velocity...it's an easy question but something is throwing me off:
An airplane wants to travel to Town A which is 650 km north of the airport. The wind velocity 125 km/h west. If the plane can fly at a velocity of 475 km/h relative to the air, determine:
a) the velocity of the plane relative to the ground and the heading (direction) that the plane must travel.
b) the time that the plane is in the air.
c) how far off course the plane would be if the pilot flew the plane ignoring the wind.
How can I even draw the two vectors..I understand that I can draw a line going 125 km/h west but the plane relative to the air 475 km/h doesn't even give the direction...can someone please help me solve it
2006-09-09 11:43:55 · 5 answers · asked by A 2 in Science & Mathematics ➔ Physics
Relative to the ground, the plane has to fly to the north.
a) Draw one vector (the wind) pointing to the west with magnitude 125 km/h.
Draw the other vector (air speed) to the north-east, such that the east pointing component is exactly 125 km/h and the magnitude of the vector is 475.
The heading relative to the north is given by theta = arcsin (125/475)= 15.257 degrees
The magnitude of the north pointing component N is now given by:
N = 475 cos(15.257) = 458.26
=> The ground-speed = 458.26km/h
b) The flight-time is now easy: 650/458.26 = 1 h 25 min 6 sec
c) If the pilot ignored the wind the flight-time would be 650/475 h and the plane would have been blown 125 x 650/475 = 171 km to the west.
2006-09-09 12:14:40 · answer #1 · answered by mitch_online_nl 3 · 0⤊ 0⤋
you can chose the direction for the 475 km/hr to counteract the wind. You can find the angle to travel at by: Angle=Arccos(125/475). Then take that angle and find the speed to the north the plane will be traveling.....speed=475*(sin(Angle from above)).
then take the 650 KM and divide by the speed you just found to find the time in the air.
for part c take the 650 and divide by 475 to find the time in air and then multiply that time by the 125 km/hr to find out how far to the west the pilot will have been blown off course.
2006-09-09 18:58:36 · answer #2 · answered by rebel_underground2003 1 · 0⤊ 0⤋
first draw the picture
the plane is going to try to go due north, but to do so, it will have to accelerate some amount to the east to overcome the westerly wind
the westerly velocity of the plane will be 125 (in order to stay going due north
the northerly velocity will be the remaining vector such that the total vector is 475 in the appropriately weighted northwesterly direction
it is difficult to explain without being able to point at the drawing
hopefully this provided some help
2006-09-09 19:13:41 · answer #3 · answered by enginerd 6 · 0⤊ 0⤋
Town A is north so the vector for the planes 475 km/h goes north.
2006-09-09 18:46:48 · answer #4 · answered by kid666_nz 3 · 0⤊ 1⤋
Normally when you talk about a west wind, it's blowing to the east. Is that what you meant here? The answers seem to interpret it the other way. Except for that, most are pretty good, I like Mitch's best.
2006-09-09 20:52:27 · answer #5 · answered by sojsail 7 · 0⤊ 0⤋