A small plane flies 40.0 km in a direction 60 degree north of east and then flie
ID: 1655979 • Letter: A
Question
A small plane flies 40.0 km in a direction 60 degree north of east and then flies 30.0 km in a direction 15 degree north of east. Use the analytical method to find the total distance the plane covers from the starting point, and the geographic direction of its displacement vector. What is its displacement vector? You fly 32.0 km in a straight line in still air in the direction 35.0 degree south of west. (a) Find the distances you would have to fly due south and then due west to arrive at the same point, (b) Find the distances you would have to fly first in a direction 45.0 degree south of west and then in a direction 45.0 degree west of north. Note these are the components of the displacement along a different set of axes-namely, the one rotated by 45 degree with respect to the axes in (a).Explanation / Answer
total distance = 40 + 30 = 70 km
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during 1st flight
r1x = 40*cos60 = 20 km
r1y = 40*sin60 = 34.64 km
during 2nd flight
r2x = 30*cos15 = 28.9 km
r2y = 30*sin15 = 7.76 km
rx = r1x + r2x = 20 + 28.9 = 48.9 km
ry = r1y + r2y = 34.64 + 7.76 = 42.4 km
geographical direction = tan^-1(ry/rx) = 40.9 North of east
displacement = sqrt(rx^2 + ry^2) = sqrt(48.9^2+42.4^2) = 64.7 km
displacement vector = 64.7 km in a direction 40.9 North of east
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17)
due to south = r*sin35 = 32*sin35 = 18.35 km
due to west = r*cos35 = 32*cos35 = 26.21 km
(b)
45 south of west
displacement = -r1*cos45i - r1*sin45 j
45 west of north
displacement = -r2*cos45i + r2*sin45 j
total displacement = -cos45*(r1 + r2)i + sin45*(r2-r1) j
but toal displacement = -32*cos35i - 32*sin35j
-32*cos35 = -cos45*(r1 + r2)
r1 + r2 = 37.1 ...........(1)
-32*sin35 = sin45*(r2 - r1)
r2 - r1 = 25.9 ............(2)
from 1 & 2
2*r2 = 63
r2 = 31.5 km
from 2
r1 = r2 - 25.9 = 31.5 - 25.9 = 5.6 km
along 45 south of west = 5.6 km
along 45 west of north = 31.5 km
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