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Answer to Question #862 in Mechanics | Relativity for duha

Question #862
A radar station locates a sinking ship at range 15.6 km and bearing 136° clockwise from north. From the same station, a rescue plane is at horizontal range 19.6 km, 160° clockwise from north, with elevation 2.10 km.
(a) Write the displacement vector from plane to ship, letting represent i east, j north, and k up.
(b) How far apart are the plane and ship?
Expert's answer
i - east, j - north, k - up:
projections for the ship:
Si = - 15.6 * cos(180°-136°) = - 11.2;
Sj = 15.6 * sin(136°) = 10.8.
Sk = 0;
Thus the vector coordinates would be (-11.2, 10.8, 0)
for the plane:
Pi = - 19.6 * cos(180° - 160°) = - 18.1;Pj = 19.6 * sin (160°) = 6.6;
Pk = 2.1
the coordinates in vector form are (-18.1, 6.6, 2.1)
The displacement vector from plane to ship equals to
S - P =(-11.2, 10.8, 0) - (-18.1, 6.6, 2.1) = (6.9, 4.2, -2.1)
The distance equals to the modulus of this vector:
sqrt(6.92+4.22+(-2.1)2) = sqrt(69.66) = 8.43 km

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Comments

Assignment Expert
19.02.14, 16:45

19.6*cos(20)=-7.99

Gary
11.02.14, 04:31

-19.6 * cos 20 does not equal to -18.1

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