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Question 253576: Hi :o)
I hope you guys can help me with this. I think I now understand how to work out a resultant vector using the coordinate system, but this question has stumped me a little. I can't get a picture of it - it seems to be 3D, is that right?
Please help...
A mine shaft goes due west 75 m from the opening at an angle of 25° below
the horizontal surface. It then becomes horizontal and turns 30° north of west
and continues for another 45 m. What is the displacement of the end of the
tunnel from the opening.
Thanks SOOOOO much :o) (Happy New Year)
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A mine shaft goes due west 75 m from the opening at an angle of 25° below
the horizontal surface. It then becomes horizontal and turns 30° north of west
and continues for another 45 m. What is the displacement of the end of the
tunnel from the opening.
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If you know how to graph in 3D,
the start point is (0,0,0)
the turning point is (-75cos(25), 0 ,-75sin(25))
the final position is (-75cos(25) ,-45,-75sin(25))
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Distance from start to finish:
d = sqrt((75cos(25))^2 + (45)^2 + (75sin(25))^2)
d = sqrt(5526.47+2025+98.53)
d = 87.46 meters
==========================
Cheers,
Stan H.
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