two airplanes leave an airport at the same time. One plane travels at 132km/h and travels due South while the other travels at 167km/h and heads due West. How far apart are the planes after 3h.
2007-07-06 04:23:49 · 6 answers · asked by Physics 101 1 in Science & Mathematics ➔ Mathematics
If 3 hours have passed, then each plane has traveled 3 times it's speed per hour. one has gone 3(132) down and the other 3(167) left. Then use pythagorean theorem to find the hypotenuse (distance)
2007-07-06 05:28:04 · answer #1 · answered by garpit c 5 · 0⤊ 0⤋
after 3 hours,
the South-bound plane has travelled
132 * 3 = 396 miles
the West-bound plane has travelled
167 * 3 = 501 miles
how far apart is the hypotenuse of the right triangle with
sides: 396 and 501
H = sqrt(396^2 + 501^2) = 638.61 miles
2007-07-06 12:15:27 · answer #2 · answered by buoisang 4 · 0⤊ 0⤋
Distance west = 3 x 167 km = 501 km
Distance south = 3 x 132 km = 396 km
d² = 501² + 396²
d² = 407817
d = 638.7
The planes are 638.7 km apart after 3 h
2007-07-06 17:58:03 · answer #3 · answered by Como 7 · 0⤊ 0⤋
Pythagoras' theorem: is x^2+y^2=z^2
where x=the distance traveled by one plane
and y = the distance traveled by the other.
This works because they are gong perpendicular to each other.
2007-07-06 11:35:59 · answer #4 · answered by monkeymobster 3 · 0⤊ 0⤋
638.6 miles away
find absolute distance then use pythagorean thm
2007-07-06 11:27:45 · answer #5 · answered by mking785 2 · 0⤊ 0⤋
--Need the latitude.
2007-07-06 13:42:27 · answer #6 · answered by Mark 6 · 0⤊ 0⤋