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March 6, 2015

March 6, 2015

Posted by **Matt, AGAIN** on Monday, February 18, 2013 at 4:40pm.

"A plane is flying 25 degrees north of west at 190 km/h and encounters a wind from 15 degrees north of east at 45 km/h. What is the planes new velocity with respect to the ground in standard position?"

Elena kindly gave me this afterward

"v=190 km/h, α= 25°,

u=45 km/h, β =15°.

V(x) =v•cos α - u•sin β =…

V(y)= v•sinα+u•sinβ=…

V=sqrt{V(x)²+V(y)²}=... "

Using this formula and inputting the number accordingly, it have me an answer of V = 171.4

The answer in the back of the book has an answer of V = 227

Is it possible the book answer is wrong?

- PHYSICS -
**Elena**, Monday, February 18, 2013 at 5:01pmI've corrected my mistake in the 1st equation

V(x) = - v•cos α +u•cosβ = - 190cos25 + 45cos15=-172.2+43.5= -128.7 km/h,

V(y)= v•sinα+u•sinβ= 190sin25 + 45sin15= 80.3 + 11.6 = 91.9 km/h,

V=sqrt{V(x)²+V(y)²}= 158.1 km/h

I believe that the answer from your book will be if

"...a wind from 15 degrees north of WEST at 45 km/h. "

The 1st equation will be

V(x) = - v•cos α - u•cosβ = - 190cos25 - 45cos15=-172.2-43.5= -216 km/h ....

- PHYSICS -
**Matt, AGAIN**, Monday, February 18, 2013 at 5:18pmThank you. So much.

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