- #1
danago
Gold Member
- 1,123
- 4
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:
[tex]
\begin{array}{l}
\overrightarrow {OB} = \left( {\begin{array}{*{20}c}
{20\cos 70} \\
{ - 20\sin 70} \\
\end{array}} \right) \\
\overrightarrow {BW} = \left( {\begin{array}{*{20}c}
0 \\
{25} \\
\end{array}} \right) \\
\end{array}
[/tex]
I then said that:
[tex]
\begin{array}{l}
\overrightarrow {BW} = \overrightarrow {BO} + \overrightarrow {OW} \\
\left( {\begin{array}{*{20}c}
0 \\
{25} \\
\end{array}} \right) = - \left( {\begin{array}{*{20}c}
{20\cos 70} \\
{ - 20\sin 70} \\
\end{array}} \right) + \overrightarrow {OW} \\
\overrightarrow {OW} = \left( {\begin{array}{*{20}c}
{20\cos 70} \\
{25 - 20\sin 70} \\
\end{array}} \right) \\
\end{array}
[/tex]
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:
[tex]
\begin{array}{l}
\overrightarrow {OB} = \left( {\begin{array}{*{20}c}
{20\cos 70} \\
{ - 20\sin 70} \\
\end{array}} \right) \\
\overrightarrow {BW} = \left( {\begin{array}{*{20}c}
0 \\
{25} \\
\end{array}} \right) \\
\end{array}
[/tex]
I then said that:
[tex]
\begin{array}{l}
\overrightarrow {BW} = \overrightarrow {BO} + \overrightarrow {OW} \\
\left( {\begin{array}{*{20}c}
0 \\
{25} \\
\end{array}} \right) = - \left( {\begin{array}{*{20}c}
{20\cos 70} \\
{ - 20\sin 70} \\
\end{array}} \right) + \overrightarrow {OW} \\
\overrightarrow {OW} = \left( {\begin{array}{*{20}c}
{20\cos 70} \\
{25 - 20\sin 70} \\
\end{array}} \right) \\
\end{array}
[/tex]
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.