relative


by EaGlE
Tags: relative
EaGlE
EaGlE is offline
#1
Sep11-04, 07:21 PM
P: 20
A canoe has a velocity of 0.420 m/s southeast relative to the earth. The canoe is on a river that is flowing 0.550 m/s east relative to the earth.

A.) Find the magnitude of the velocity of the canoe relative to the river.

B.) Find the direction of the velocity of the canoe relative to the river. Express your answer as an angle measured south of west.


ok first thing i did was drew a picture, cause i dont get the question.

to find the unknown side(vector Y) i did

0.550^2 + Y^2 = 0.420^2

Y = 0.355m/s

i thought that vector Y was unknown because it gave the southeast side, which is the longest side right? and it also gave the east side, which is vectory X.


and for B.) arctan(0.355/0.550) = 32.84 degrees

well that's all i did so far, i dont even know if either of them is correct. i really doubt that any of them are correct, because it seemed to easy.
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HallsofIvy
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#2
Sep11-04, 08:53 PM
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While you say "vector", you are treating Y as a number. In other words, you are assuming that the motion of the canoe "relative to the river" is due south and, from the wording of the question, I don't believe that's true. In fact, using the equation you give, 0.550^2 + Y^2 = 0.420^2, you get Y^2= -.1261 which is impossible.
Suppose the motion of the canoe, relative to the river, is v m/s at angle θ east of n. Drawing your picture (always a good idea) you should have a triangle (NOT a right triangle) with one angle 45 degrees and the other θ degrees (the third is, of course 180-45- &theta) and two lengths .420 and .55. The unknown side is opposite the known 45 degree angle so you can use the "cosine law" to find its length and then use the "sine law" to find θ.
EaGlE
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#3
Sep11-04, 10:21 PM
P: 20
Quote Quote by HallsofIvy
While you say "vector", you are treating Y as a number. In other words, you are assuming that the motion of the canoe "relative to the river" is due south and, from the wording of the question, I don't believe that's true. In fact, using the equation you give, 0.550^2 + Y^2 = 0.420^2, you get Y^2= -.1261 which is impossible.
Suppose the motion of the canoe, relative to the river, is v m/s at angle θ east of n. Drawing your picture (always a good idea) you should have a triangle (NOT a right triangle) with one angle 45 degrees and the other θ degrees (the third is, of course 180-45- &theta) and two lengths .420 and .55. The unknown side is opposite the known 45 degree angle so you can use the "cosine law" to find its length and then use the "sine law" to find θ.

how did you get 45 degrees?

jdstokes
jdstokes is offline
#4
Sep11-04, 11:57 PM
P: 527

relative


Quote Quote by EaGlE
A canoe has a velocity of 0.420 m/s southeast relative to the earth. The canoe is on a river that is flowing 0.550 m/s east relative to the earth.

A.) Find the magnitude of the velocity of the canoe relative to the river.

B.) Find the direction of the velocity of the canoe relative to the river. Express your answer as an angle measured south of west.


ok first thing i did was drew a picture, cause i dont get the question.

to find the unknown side(vector Y) i did

0.550^2 + Y^2 = 0.420^2

Y = 0.355m/s

i thought that vector Y was unknown because it gave the southeast side, which is the longest side right? and it also gave the east side, which is vectory X.


and for B.) arctan(0.355/0.550) = 32.84 degrees

well that's all i did so far, i dont even know if either of them is correct. i really doubt that any of them are correct, because it seemed to easy.
[latex]\begin{align}
\intertext{The vector equation}
\mathbf{v}_\mathrm{CR} &=\mathbf{v}_\mathrm{CE}-\mathbf{v}_\mathrm{RE}\\
\intertext{gives the velocity of the canoe relative to the river. You can decompose the right hand side of (1) into components, with $\mathbf{i}$ and $\mathbf{j}$ denoting the unit vectors pointing east and north, respectively}
\mathbf{v}_\mathrm{CR}& = \frac{v_\mathrm{CE}}{\sqrt{2}}\mathbf{i}-\frac{v_\mathrm{CE}}{\sqrt{2}}\mathbf{j}- v_\mathrm{RE}\mathbf{i}.
\end{align}[/latex]
EaGlE
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#5
Sep12-04, 02:04 AM
P: 20
Quote Quote by jdstokes
[latex]\begin{align}
\intertext{The vector equation}
\mathbf{v}_\mathrm{CR} &=\mathbf{v}_\mathrm{CE}-\mathbf{v}_\mathrm{RE}\\
\intertext{gives the velocity of the canoe relative to the river. You can decompose the right hand side of (1) into components, with $\mathbf{i}$ and $\mathbf{j}$ denoting the unit vectors pointing east and north, respectively}
\mathbf{v}_\mathrm{CR}& = \frac{v_\mathrm{CE}}{\sqrt{2}}\mathbf{i}-\frac{v_\mathrm{CE}}{\sqrt{2}}\mathbf{j}- v_\mathrm{RE}\mathbf{i}.
\end{align}[/latex]

[latex]\mathbf{v}_\mathrm{CR}& = \frac{v_\mathrm{CE}}{\sqrt{2}}\mathbf{i}-\frac{v_\mathrm{CE}}{\sqrt{2}}\mathbf{j}- v_\mathrm{RE}\mathbf{i}.
\end{align}[/latex]

^hmmm, why do i need to sqrt it by 2? and why is vector i and j the same?

Vc/r = Vc/e - Vr/e
vc/r = 0.420m/s - 0.55m/s

im sorry, but i am soo confused. and can you tell me how you got 45 degrees

ok i have my picture attached that i draw, what can i tell from the picture?

and how do you tell which vector velocity is i and which is j?
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HallsofIvy
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#6
Sep12-04, 09:30 AM
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Draw a picture.
"A canoe has a velocity of 0.420 m/s southeast relative to the earth. " while the river is running due east. Don't you see that that means that the path of the canoe is at 45 degrees to the river?

As for jdstokes' [itex]\frac{1}{\sqrt{2}}[/itex], that is the ratio of the leg of 45-45-90 right triangle to the hypotenuse. Use the Pythagorean formula: a2+ b2= c2. In a 45-45-90 right triangle, the legs are of equal length, call it x. We have x2+ x2= 2x2= c2 so [itex](\frac{x}{c})^2= \frac{1}{2}[/itex] and [itex]\frac{x}{c}= \frac{1}{\sqrt{2}}[/itex].
Or you could use trigonometry: sin(45)= cos(45)= [itex]\frac{1}{\sqrt{2}}[/itex].
EaGlE
EaGlE is offline
#7
Sep13-04, 01:47 AM
P: 20
Don't you see that that means that the path of the canoe is at 45 degrees to the river
i see how you got 45 degrees now, i been drawing the wrong picture !


As for jdstokes' [itex]\frac{1}{\sqrt{2}}[/itex], that is the ratio of the leg of 45-45-90 right triangle to the hypotenuse. Use the Pythagorean formula: a2+ b2= c2. In a 45-45-90 right triangle, the legs are of equal length, call it x. We have x2+ x2= 2x2= c2 so [itex](\frac{x}{c})^2= \frac{1}{2}[/itex] and [itex]\frac{x}{c}= \frac{1}{\sqrt{2}}[/itex].
Or you could use trigonometry: sin(45)= cos(45)= [itex]\frac{1}{\sqrt{2}}[/itex].
yea im more comfortable with using trigonometry, makes more sense now.
A.) Find the magnitude of the velocity of the canoe relative to the river.
i got -.25301(i)^2 - .29698(j)^2 = C^2 (not sure if both should be negative)
c= .3980145 <-- it's the correct answer, so there's no need to check this answer.

ok for B.) Find the direction of the velocity of the canoe relative to the river. Express your answer as an angle measured south of west. isnt it just arctan(y/x).. or in this case arctan(j/i)... so... arctan(-29698/-.25301)= 89.9 Degrees. but it wants it wants the degrees in south of west. which direction is 89.9 degrees pointing? and how would i switch it to south of west.

i know my directions(North,East,South,West), i just dont know how the initial direction and how to convert it to something else.


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