Mechanics
"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?
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?
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?
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.
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?
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?
Original post by Muttley79
Is this the whole question?
Yep
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.
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 ...
Speaking of Vectors, Anyone have any idea on these questions?
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.
Original post by marinacalder
Speaking of Vectors, Anyone have any idea on these questions?
How far have you got?
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
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?
Original post by Muttley79
How without more information? I think the question is incomplete ...
Sorry it was the 2015 question, I typed it wrong
Original post by Lucofthewoods
Sorry it was the 2015 question, I typed it wrong
Have you worked it out now?
Original post by Muttley79
Have you worked it out now?
Nope, struggling with mechanics so much
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.
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.
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!
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 = Qsin@, Horiz: Rx = 0 => P = 45 - Q = Qcos@ => 45 = Q(1 + cos@)
So 3 = (1 + cos@)/sin@
=> 3sin@ = 1 + cos@
=> 3sin@ - cos@ = 1
3sin@ - cos@ = Rsin(@ - a) = [Rcos(a)]sin@ - [Rsin(a)]cos@. Equate co-effs of sin@, cos@:
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 - 2@/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 = Qsin@, Horiz: Rx = P - Qsin@
R^2 = (P - Qsin@)^2 + (Qsin@)^2
Eq 1: Sub P = 2Rt3, Q = 2. Eq2: Sub P = 2Rt3, Q = 4.
Equate: (2Rt3 - 2cos@)^2 + (2sin@)^2 = (2Rt3 - 4cos@)^2 + (4sin@)^2
=> 12 - 8Rt3.cos@ + 4[(cos@)^2 + (sin@)^2] = 12 - 16Rt3.cos@ + 16[(cos@)^2 + (sin@)^2]
=> 8Rt3.cos@ = 12
=> cos@ = (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
(edited 5 years ago)
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