Mixing
application of vectors: sailboat floats in a current
$begingroup$
A sailboat floats in a current that flows due east at 1 meter per second. Due to a wind the boats actual speed relative to the shore is $({sqrt 3})$ meters per second in a direction 30 degrees North of East. Find the speed and direction of the wind.
So far I have found the speed of the wind by using the formula for the resultant vector and got the speed to be 1 meter per second. Now how do I go about finding the direction of the wind? Can someone provide a step by step explanation. I don't understand why the wind would be East of North?
vectors
asked Feb 1 '16 at 15:20
LilLil
95842545
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add a comment |
$begingroup$
A sailboat floats in a current that flows due east at 1 meter per second. Due to a wind the boats actual speed relative to the shore is $({sqrt 3})$ meters per second in a direction 30 degrees North of East. Find the speed and direction of the wind.
So far I have found the speed of the wind by using the formula for the resultant vector and got the speed to be 1 meter per second. Now how do I go about finding the direction of the wind? Can someone provide a step by step explanation. I don't understand why the wind would be East of North?
vectors
asked Feb 1 '16 at 15:20
LilLil
95842545
$endgroup$
add a comment |
$begingroup$
A sailboat floats in a current that flows due east at 1 meter per second. Due to a wind the boats actual speed relative to the shore is $({sqrt 3})$ meters per second in a direction 30 degrees North of East. Find the speed and direction of the wind.
So far I have found the speed of the wind by using the formula for the resultant vector and got the speed to be 1 meter per second. Now how do I go about finding the direction of the wind? Can someone provide a step by step explanation. I don't understand why the wind would be East of North?
vectors
asked Feb 1 '16 at 15:20
LilLil
95842545
$endgroup$
A sailboat floats in a current that flows due east at 1 meter per second. Due to a wind the boats actual speed relative to the shore is $({sqrt 3})$ meters per second in a direction 30 degrees North of East. Find the speed and direction of the wind.
So far I have found the speed of the wind by using the formula for the resultant vector and got the speed to be 1 meter per second. Now how do I go about finding the direction of the wind? Can someone provide a step by step explanation. I don't understand why the wind would be East of North?
vectors
vectors
asked Feb 1 '16 at 15:20
LilLil
95842545
asked Feb 1 '16 at 15:20
LilLil
95842545
edited Feb 1 '16 at 15:21
Lil
asked Feb 1 '16 at 15:20
LilLil
95842545
asked Feb 1 '16 at 15:20
LilLil
95842545
asked Feb 1 '16 at 15:20
LilLil
95842545
95842545
add a comment |
add a comment |
1 Answer
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$begingroup$
The Actual velocity $(vec{v_r})$ rel to shore will be resultant of sailboat velocity $(vec{v_b})$ and wind velocity $(vec{v_w})$.
Figure 1:
Figure 2:
$theta = 180 -30 = 150^circ$
$$begin{align} vec{v_r} &= vec{v_w}+vec{v_b} \ \
implies vec{v_w} &= vec{v_r}-vec{v_b}
end{align}$$
$$begin{align}|vec{v_w}| &= sqrt{|vec{v_r}|^2 + |vec{v_b}|^2 +2|vec{v_r}||vec{v_b}| costheta} \ \
&=sqrt{3 + 1 - 2dfrac{3}{2}} \ &=1
end{align}$$
$$begin{align} tanphi &= dfrac{|vec{v_b}|sintheta}{|vec{v_r}|+|vec{v_b}|costheta} \ \
&= dfrac{1/2}{sqrt{3}-frac{sqrt{3}}{2}} \
&= frac{1}{sqrt{3}}\ \
implies phi &= 30^circ
end{align}$$
Therefore direction of wind is $30^circ$ east of north, or $60^circ$ north of east.
answered Feb 1 '16 at 17:19
Max PayneMax Payne
2,552825
$endgroup$
$begingroup$
after getting the value 30 how did you know it automatically had to be east of north though?
$endgroup$
– Lil
Feb 1 '16 at 18:23
$begingroup$
@Lil you can either use law of cosines, or the angle formula of $tanphi$. In that formula, $phi$ is the angle made from first vector, first in the sense the one which comes first in the order of addition. So $phi$ is angle made between $vec{v_w}$ and $vec{v_r}$ and not the angle made between $vec{v_w}$ and $-vec{v_b}$
$endgroup$
– Max Payne
Feb 2 '16 at 4:34
$begingroup$
and $vec{v_r}$ is $30^circ$ north of east, and $vec{v_w}$ is $30^circ$ north of $vec{v_r}$, So $vec{v_w}$ is $60^circ$ north of east, or $30^circ$ east of north.
$endgroup$
– Max Payne
Feb 2 '16 at 4:39
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
The Actual velocity $(vec{v_r})$ rel to shore will be resultant of sailboat velocity $(vec{v_b})$ and wind velocity $(vec{v_w})$.
Figure 1:
Figure 2:
$theta = 180 -30 = 150^circ$
$$begin{align} vec{v_r} &= vec{v_w}+vec{v_b} \ \
implies vec{v_w} &= vec{v_r}-vec{v_b}
end{align}$$
$$begin{align}|vec{v_w}| &= sqrt{|vec{v_r}|^2 + |vec{v_b}|^2 +2|vec{v_r}||vec{v_b}| costheta} \ \
&=sqrt{3 + 1 - 2dfrac{3}{2}} \ &=1
end{align}$$
$$begin{align} tanphi &= dfrac{|vec{v_b}|sintheta}{|vec{v_r}|+|vec{v_b}|costheta} \ \
&= dfrac{1/2}{sqrt{3}-frac{sqrt{3}}{2}} \
&= frac{1}{sqrt{3}}\ \
implies phi &= 30^circ
end{align}$$
Therefore direction of wind is $30^circ$ east of north, or $60^circ$ north of east.
answered Feb 1 '16 at 17:19
Max PayneMax Payne
2,552825
$endgroup$
$begingroup$
after getting the value 30 how did you know it automatically had to be east of north though?
$endgroup$
– Lil
Feb 1 '16 at 18:23
$begingroup$
@Lil you can either use law of cosines, or the angle formula of $tanphi$. In that formula, $phi$ is the angle made from first vector, first in the sense the one which comes first in the order of addition. So $phi$ is angle made between $vec{v_w}$ and $vec{v_r}$ and not the angle made between $vec{v_w}$ and $-vec{v_b}$
$endgroup$
– Max Payne
Feb 2 '16 at 4:34
$begingroup$
and $vec{v_r}$ is $30^circ$ north of east, and $vec{v_w}$ is $30^circ$ north of $vec{v_r}$, So $vec{v_w}$ is $60^circ$ north of east, or $30^circ$ east of north.
$endgroup$
– Max Payne
Feb 2 '16 at 4:39
add a comment |
$begingroup$
The Actual velocity $(vec{v_r})$ rel to shore will be resultant of sailboat velocity $(vec{v_b})$ and wind velocity $(vec{v_w})$.
Figure 1:
Figure 2:
$theta = 180 -30 = 150^circ$
$$begin{align} vec{v_r} &= vec{v_w}+vec{v_b} \ \
implies vec{v_w} &= vec{v_r}-vec{v_b}
end{align}$$
$$begin{align}|vec{v_w}| &= sqrt{|vec{v_r}|^2 + |vec{v_b}|^2 +2|vec{v_r}||vec{v_b}| costheta} \ \
&=sqrt{3 + 1 - 2dfrac{3}{2}} \ &=1
end{align}$$
$$begin{align} tanphi &= dfrac{|vec{v_b}|sintheta}{|vec{v_r}|+|vec{v_b}|costheta} \ \
&= dfrac{1/2}{sqrt{3}-frac{sqrt{3}}{2}} \
&= frac{1}{sqrt{3}}\ \
implies phi &= 30^circ
end{align}$$
Therefore direction of wind is $30^circ$ east of north, or $60^circ$ north of east.
answered Feb 1 '16 at 17:19
Max PayneMax Payne
2,552825
$endgroup$
$begingroup$
after getting the value 30 how did you know it automatically had to be east of north though?
$endgroup$
– Lil
Feb 1 '16 at 18:23
$begingroup$
@Lil you can either use law of cosines, or the angle formula of $tanphi$. In that formula, $phi$ is the angle made from first vector, first in the sense the one which comes first in the order of addition. So $phi$ is angle made between $vec{v_w}$ and $vec{v_r}$ and not the angle made between $vec{v_w}$ and $-vec{v_b}$
$endgroup$
– Max Payne
Feb 2 '16 at 4:34
$begingroup$
and $vec{v_r}$ is $30^circ$ north of east, and $vec{v_w}$ is $30^circ$ north of $vec{v_r}$, So $vec{v_w}$ is $60^circ$ north of east, or $30^circ$ east of north.
$endgroup$
– Max Payne
Feb 2 '16 at 4:39
add a comment |
$begingroup$
The Actual velocity $(vec{v_r})$ rel to shore will be resultant of sailboat velocity $(vec{v_b})$ and wind velocity $(vec{v_w})$.
Figure 1:
Figure 2:
$theta = 180 -30 = 150^circ$
$$begin{align} vec{v_r} &= vec{v_w}+vec{v_b} \ \
implies vec{v_w} &= vec{v_r}-vec{v_b}
end{align}$$
$$begin{align}|vec{v_w}| &= sqrt{|vec{v_r}|^2 + |vec{v_b}|^2 +2|vec{v_r}||vec{v_b}| costheta} \ \
&=sqrt{3 + 1 - 2dfrac{3}{2}} \ &=1
end{align}$$
$$begin{align} tanphi &= dfrac{|vec{v_b}|sintheta}{|vec{v_r}|+|vec{v_b}|costheta} \ \
&= dfrac{1/2}{sqrt{3}-frac{sqrt{3}}{2}} \
&= frac{1}{sqrt{3}}\ \
implies phi &= 30^circ
end{align}$$
Therefore direction of wind is $30^circ$ east of north, or $60^circ$ north of east.
answered Feb 1 '16 at 17:19
Max PayneMax Payne
2,552825
$endgroup$
The Actual velocity $(vec{v_r})$ rel to shore will be resultant of sailboat velocity $(vec{v_b})$ and wind velocity $(vec{v_w})$.
Figure 1:
Figure 2:
$theta = 180 -30 = 150^circ$
$$begin{align} vec{v_r} &= vec{v_w}+vec{v_b} \ \
implies vec{v_w} &= vec{v_r}-vec{v_b}
end{align}$$
$$begin{align}|vec{v_w}| &= sqrt{|vec{v_r}|^2 + |vec{v_b}|^2 +2|vec{v_r}||vec{v_b}| costheta} \ \
&=sqrt{3 + 1 - 2dfrac{3}{2}} \ &=1
end{align}$$
$$begin{align} tanphi &= dfrac{|vec{v_b}|sintheta}{|vec{v_r}|+|vec{v_b}|costheta} \ \
&= dfrac{1/2}{sqrt{3}-frac{sqrt{3}}{2}} \
&= frac{1}{sqrt{3}}\ \
implies phi &= 30^circ
end{align}$$
Therefore direction of wind is $30^circ$ east of north, or $60^circ$ north of east.
answered Feb 1 '16 at 17:19
Max PayneMax Payne
2,552825
answered Feb 1 '16 at 17:19
Max PayneMax Payne
2,552825
answered Feb 1 '16 at 17:19
Max PayneMax Payne
2,552825
answered Feb 1 '16 at 17:19
Max PayneMax Payne
2,552825
2,552825
$begingroup$
after getting the value 30 how did you know it automatically had to be east of north though?
$endgroup$
– Lil
Feb 1 '16 at 18:23
$begingroup$
@Lil you can either use law of cosines, or the angle formula of $tanphi$. In that formula, $phi$ is the angle made from first vector, first in the sense the one which comes first in the order of addition. So $phi$ is angle made between $vec{v_w}$ and $vec{v_r}$ and not the angle made between $vec{v_w}$ and $-vec{v_b}$
$endgroup$
– Max Payne
Feb 2 '16 at 4:34
$begingroup$
and $vec{v_r}$ is $30^circ$ north of east, and $vec{v_w}$ is $30^circ$ north of $vec{v_r}$, So $vec{v_w}$ is $60^circ$ north of east, or $30^circ$ east of north.
$endgroup$
– Max Payne
Feb 2 '16 at 4:39
add a comment |
$begingroup$
after getting the value 30 how did you know it automatically had to be east of north though?
$endgroup$
– Lil
Feb 1 '16 at 18:23
$begingroup$
@Lil you can either use law of cosines, or the angle formula of $tanphi$. In that formula, $phi$ is the angle made from first vector, first in the sense the one which comes first in the order of addition. So $phi$ is angle made between $vec{v_w}$ and $vec{v_r}$ and not the angle made between $vec{v_w}$ and $-vec{v_b}$
$endgroup$
– Max Payne
Feb 2 '16 at 4:34
$begingroup$
and $vec{v_r}$ is $30^circ$ north of east, and $vec{v_w}$ is $30^circ$ north of $vec{v_r}$, So $vec{v_w}$ is $60^circ$ north of east, or $30^circ$ east of north.
$endgroup$
– Max Payne
Feb 2 '16 at 4:39
$begingroup$
after getting the value 30 how did you know it automatically had to be east of north though?
$endgroup$
– Lil
Feb 1 '16 at 18:23
$begingroup$
after getting the value 30 how did you know it automatically had to be east of north though?
$endgroup$
– Lil
Feb 1 '16 at 18:23
$begingroup$
@Lil you can either use law of cosines, or the angle formula of $tanphi$. In that formula, $phi$ is the angle made from first vector, first in the sense the one which comes first in the order of addition. So $phi$ is angle made between $vec{v_w}$ and $vec{v_r}$ and not the angle made between $vec{v_w}$ and $-vec{v_b}$
$endgroup$
– Max Payne
Feb 2 '16 at 4:34
$begingroup$
@Lil you can either use law of cosines, or the angle formula of $tanphi$. In that formula, $phi$ is the angle made from first vector, first in the sense the one which comes first in the order of addition. So $phi$ is angle made between $vec{v_w}$ and $vec{v_r}$ and not the angle made between $vec{v_w}$ and $-vec{v_b}$
$endgroup$
– Max Payne
Feb 2 '16 at 4:34
$begingroup$
and $vec{v_r}$ is $30^circ$ north of east, and $vec{v_w}$ is $30^circ$ north of $vec{v_r}$, So $vec{v_w}$ is $60^circ$ north of east, or $30^circ$ east of north.
$endgroup$
– Max Payne
Feb 2 '16 at 4:39
$begingroup$
and $vec{v_r}$ is $30^circ$ north of east, and $vec{v_w}$ is $30^circ$ north of $vec{v_r}$, So $vec{v_w}$ is $60^circ$ north of east, or $30^circ$ east of north.
$endgroup$
– Max Payne
Feb 2 '16 at 4:39
add a comment |
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