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Two airplanes leave an airport at the same time. The velocity of the first airplane is 730 m/h at a heading of 65.3 â—¦ . The velocity of the second is 590 m/h at a heading of 102â—¦ . How far apart are they after 1.7 h? Answer in units of m.

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Answer:

Plane will 741.6959 m apart after 1.7 hour                    

Explanation:

We have given time = 1.7 hr

So if we draw the vectors of a 2d graph we see that the difference in angles is   = [tex]102^{\circ}-65.3^{\circ}=36.7^{\circ}[/tex]

Speed of first plane  = 730 m/h

So distance traveled by first plane = 730×1.7 = 1241 m

Speed of second plane = 590 m/hr

So distance traveled by second plane = 590×1.7 = 1003 m

We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 58.6.

Using the law of cosine, [tex]r^2[/tex] representing the distance between the planes, we see that:

[tex]r^2=1241^2+1003^2-2\times 1003\times 1241cos(36.7)=550112.8295[/tex]

r = 741.6959 m

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