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Old Oct 6th 2009, 04:48 PM   #1
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help needed... velocity, and magnitude

A wind is blowing directly from east to west. The pilot of a small plane finds that if he points the nose of the plane 32.8 ° north of east, his velocity with respect to the ground is in the direction 53.6 ° north of east. The speed of the plane with respect to the air is 133 m/s. Taking North to be the y-direction and East to be the x-direction, what is the y-component of plane's velocity with respect to the ground and what is the magnitude of the plane's velocity with respect to the ground?

i thought it should be:
133sin53.6= 107.05 m/s beause 53.6 degrees is that respect to the ground... but it is wrong answer... hmm...

anyone can help me out a little? thanks !!
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Old Oct 6th 2009, 11:44 PM   #2
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are the answers 72m/sec (y component) and 89.4m/sec(plane velocity wrt ground)? if they are not, please reply. i will think again.
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Old Oct 7th 2009, 09:05 AM   #3
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the answer for the x-component is correct
but on y-component, significant digits are wrong.

would you mind if you explain me how you solved please...? and correct answer within significant digits?
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Old Oct 7th 2009, 09:13 AM   #4
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aha !!
i got it after !!

133 sin 32.8 for y component

but how did you get the x component?
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Old Oct 8th 2009, 07:12 AM   #5
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sorry for late reply.(my exam is at door)
perhaps i misunderstood the question , but it said clearly to find y component velocity and the magnitude of plane velocity(NOT the x component)
here is what i did (note v(p,a)=vel of plane wrt air.v(p,g)=vel of plane wrt ground.v(a,g)=vel of air wrt ground )
v(p,a)=133(given) so v(p,a)=v(p,g)-v(a,g) let v(a,g)=-x i(i stand for unit vector along x and wind flows from east to west, so minus sign)
so v(p,g)=v(p,a)-x i=133cos32.8 i +133 sin32.8 j-xi=(133cos32.8-x)i+133sin32.8 j
i then used tan 53.6=(133sin32.8)/(133cos32.8 -x) from which i solved x which came 58.69. put it in v(p,g) above and we get the complete vector. now you find out what you need.
i apologize for the above used abbreviations.
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Old Oct 8th 2009, 12:09 PM   #6
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thank you for the detail explanation
i totally got it now... hehe
thank you so much
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