danago
Gold Member
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Hey. I've done this question, but my answer is only partly correct according to the answer book. Heres the question:
"To a bird flying at 20km/h on a bearing of 160 degrees, the wind seems to be coming from the south at 25km/h. Find the true velocity of the wind"
ok. I said that B is the velocity of the bird, W is the velocity of the wind, and O is the origin. I then found that:
<br /> \begin{array}{l}<br /> \overrightarrow {OB} = \left( {\begin{array}{*{20}c}<br /> {20\cos 70} \\<br /> { - 20\sin 70} \\<br /> \end{array}} \right) \\ <br /> \overrightarrow {BW} = \left( {\begin{array}{*{20}c}<br /> 0 \\<br /> {25} \\<br /> \end{array}} \right) \\ <br /> \end{array}<br />
I then said that:
<br /> \begin{array}{l}<br /> \overrightarrow {BW} = \overrightarrow {BO} + \overrightarrow {OW} \\ <br /> \left( {\begin{array}{*{20}c}<br /> 0 \\<br /> {25} \\<br /> \end{array}} \right) = - \left( {\begin{array}{*{20}c}<br /> {20\cos 70} \\<br /> { - 20\sin 70} \\<br /> \end{array}} \right) + \overrightarrow {OW} \\ <br /> \overrightarrow {OW} = \left( {\begin{array}{*{20}c}<br /> {20\cos 70} \\<br /> {25 - 20\sin 70} \\<br /> \end{array}} \right) \\ <br /> \end{array}<br />
I then found the answer to be 9.24km/h on a bearing of 48 degrees, but the answer says 9.24hm/h on a bearing of 228 degrees. Have i gone wrong somewhere? is the book wrong? any help is greatly appreciated.
Thanks,
Dan.
"To a bird flying at 20km/h on a bearing of 160 degrees, the wind seems to be coming from the south at 25km/h. Find the true velocity of the wind"
ok. I said that B is the velocity of the bird, W is the velocity of the wind, and O is the origin. I then found that:
<br /> \begin{array}{l}<br /> \overrightarrow {OB} = \left( {\begin{array}{*{20}c}<br /> {20\cos 70} \\<br /> { - 20\sin 70} \\<br /> \end{array}} \right) \\ <br /> \overrightarrow {BW} = \left( {\begin{array}{*{20}c}<br /> 0 \\<br /> {25} \\<br /> \end{array}} \right) \\ <br /> \end{array}<br />
I then said that:
<br /> \begin{array}{l}<br /> \overrightarrow {BW} = \overrightarrow {BO} + \overrightarrow {OW} \\ <br /> \left( {\begin{array}{*{20}c}<br /> 0 \\<br /> {25} \\<br /> \end{array}} \right) = - \left( {\begin{array}{*{20}c}<br /> {20\cos 70} \\<br /> { - 20\sin 70} \\<br /> \end{array}} \right) + \overrightarrow {OW} \\ <br /> \overrightarrow {OW} = \left( {\begin{array}{*{20}c}<br /> {20\cos 70} \\<br /> {25 - 20\sin 70} \\<br /> \end{array}} \right) \\ <br /> \end{array}<br />
I then found the answer to be 9.24km/h on a bearing of 48 degrees, but the answer says 9.24hm/h on a bearing of 228 degrees. Have i gone wrong somewhere? is the book wrong? any help is greatly appreciated.
Thanks,
Dan.