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    "A yacht is sailing through water that is flowing due west at 2ms-1. Velocity of yacht relative to water is 6ms-1. The yacht has a resultant velocity of V ms-1 on a bearing of theta degrees"

    I worked out V using Pythagoras 6.32ms-1 but how do you find the bearing? could someone explain it well, possibly with a diagram?
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    (Original post by Lucofthewoods)
    "A yacht is sailing through water that is flowing due west at 2ms-1. Velocity of yacht relative to water is 6ms-1. The yacht has a resultant velocity of V ms-1 on a bearing of theta degrees"

    I worked out V using Pythagoras 6.32ms-1 but how do you find the bearing? could someone explain it well, possibly with a diagram?
    If you draw the triangle of velocities, you should be able to use basic trig to get the angle.
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    (Original post by Lucofthewoods)
    "A yacht is sailing through water that is flowing due west at 2ms-1. Velocity of yacht relative to water is 6ms-1. The yacht has a resultant velocity of V ms-1 on a bearing of theta degrees"

    I worked out V using Pythagoras 6.32ms-1 but how do you find the bearing? could someone explain it well, possibly with a diagram?
    Is this the whole question?
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    (Original post by Muttley79)
    Is this the whole question?
    Yep
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    (Original post by Lucofthewoods)
    Yep
    Are you sure the direction of the yacht is not given?

    It's very similar to Q2 of AQA June 2015 ...


    A yacht is sailing through water that is flowing due west at 2ms^-1. The velocity of the yacht relative to the water is 6ms^-1 due south.
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    (Original post by Zacken)
    If you draw the triangle of velocities, you should be able to use basic trig to get the angle.
    How without more information? I think the question is incomplete ...
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    Speaking of Vectors, Anyone have any idea on these questions?Name:  IMG_2074.jpg
Views: 59
Size:  500.5 KB
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    (Original post by Muttley79)
    How without more information? I think the question is incomplete ...
    Ah yeah, I didn't look too closely, at a second glance, I agree with you.
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    (Original post by marinacalder)
    Speaking of Vectors, Anyone have any idea on these questions?Name:  IMG_2074.jpg
Views: 59
Size:  500.5 KB
    How far have you got?
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    (Original post by Muttley79)
    How far have you got?
    Iv'e tried various things on all 3- is the scalar product involved? I've tried drawing and using trig/ pythagoras... nothing's working
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    (Original post by marinacalder)
    Iv'e tried various things on all 3- is the scalar product involved? I've tried drawing and using trig/ pythagoras... nothing's working
    Can you post anything you've written?
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    Name:  IMG_2090.jpg
Views: 71
Size:  491.1 KB
    (Original post by Muttley79)
    Can you post anything you've written?
    Attached Images
     
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    (Original post by Muttley79)
    How without more information? I think the question is incomplete ...
    Sorry it was the 2015 question, I typed it wrong
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    Dunno about vectors but applying the matrix \begin{pmatrix}0 & 1 \\-1 & 0\end{pmatrix} to your attachments would make them much easier to read...
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    (Original post by Lucofthewoods)
    Sorry it was the 2015 question, I typed it wrong
    Have you worked it out now?
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    (Original post by Muttley79)
    Have you worked it out now?
    Nope, struggling with mechanics so much
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    (Original post by Lucofthewoods)
    Nope, struggling with mechanics so much
    Sorry to hear that

    A lot of students struggle with vectors so keep going.

    "A yacht is sailing through water that is flowing due west at 2ms-1. Velocity of yacht relative to water is 6ms-1 due south. The yacht has a resultant velocity of V ms-1 on a bearing of theta degrees"

    You worked out the velocity from Pythagoras - you needed a right angled traingle for that?
    Presumably you had a line of length 2 with an arrow pointing west representing the river then a line of length 6 pointing south representing the yacht drawn from the end of the previous arrow? Then you can draw the resultant velocity and direction..

    Use the same triangle to get the angle between the water and the yacht's direction then you need to change it into a bearing.
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    (Original post by Muttley79)
    Sorry to hear that

    A lot of students struggle with vectors so keep going.

    "A yacht is sailing through water that is flowing due west at 2ms-1. Velocity of yacht relative to water is 6ms-1 due south. The yacht has a resultant velocity of V ms-1 on a bearing of theta degrees"

    You worked out the velocity from Pythagoras - you needed a right angled traingle for that?
    Presumably you had a line of length 2 with an arrow pointing west representing the river then a line of length 6 pointing south representing the yacht drawn from the end of the previous arrow? Then you can draw the resultant velocity and direction..

    Use the same triangle to get the angle between the water and the yacht's direction then you need to change it into a bearing.
    Ahhh thank you that makes so much more sense!
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    (Original post by marinacalder)
    ...
    6) P & Q make an obtuse angle, R is in-between at 90° to P. Let P lie horiz, so R is vert and Q at @ to horiz.

    Resolve Vert: Rv = R = 15 = [email protected], Horiz: Rx = 0 => P = 45 - Q = [email protected] => 45 = Q(1 + [email protected])

    So 3 = (1 + [email protected])/[email protected]
    => [email protected] = 1 + [email protected]
    => [email protected] - [email protected] = 1

    [email protected] - [email protected] = Rsin(@ - a) = [Rcos(a)][email protected] - [Rsin(a)][email protected] Equate co-effs of [email protected], [email protected]:

    Rcos(a) = 3, Rsin(a) = 1
    R^2 = 3^2 + 1^2 = 10 => R = Rt10
    tan(a) = 1/3 => a = 18.435°

    So Rt10.sin(@ - 18.435°) = 1
    => @ - 18.435° = sin^-1(1/Rt10) = 18.435°
    => @ = 36.9°, Theta = 180° - @ = 143.1°

    Q = 15/sin(36.9°) = 25N, P = 45 - Q = 20N

    7) F // (24i + 7j) => F = a(24i + 7j), a = scalar multiplier
    |24i + 7j| = 25 = |F|/2 => a = 2 => F = 48i + 14j

    F1 = a(i - 2j), F2 = b(4i + 3j)
    @ = a.Rt5, B = b.Rt25 = 5b

    So F1 = (@/Rt5)(i - 2j), F2 = (B/5)(4i + 3j)
    => (@/Rt5)(i - 2j) + (B/5)(4i + 3j) = 48i + 14j

    i co-effs: @/Rt5 + 4B/5 = 48 => @.Rt5 + 4B = 240
    j co-effs: 3B/5 - [email protected]/Rt5 = 14 => 3B - [email protected] = 70

    Solve simult: @ = 8.Rt5, B = 50

    8) Same set-up as Q6. Q adds to Rv but subtracts from Rx (from P). So 2 Q values can result in the same R.

    Resolve Vert: Rv = [email protected], Horiz: Rx = P - [email protected]
    R^2 = (P - [email protected])^2 + ([email protected])^2
    Eq 1: Sub P = 2Rt3, Q = 2. Eq2: Sub P = 2Rt3, Q = 4.

    Equate: (2Rt3 - [email protected])^2 + ([email protected])^2 = (2Rt3 - [email protected])^2 + ([email protected])^2

    => 12 - [email protected] + 4[([email protected])^2 + ([email protected])^2] = 12 - [email protected] + 16[([email protected])^2 + ([email protected])^2]

    => [email protected] = 12
    => [email protected] = (Rt3)/2
    => @ = 30°, Theta = 180° - @ = 150°

    For R, sub Q = 2 into prev exp for R^2:
    R^2 = (2Rt3 - 2cos30°)^2 + (2sin30°)^2
    => R^2 = (Rt3)^2 + 1^2 = 4
    => R = 2
 
 
 

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