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Need help on how to draw resultant velocity triangles!

Hi, so MM1B exam for AQA is on Tuesday. And basically, i dont understand how to properly draw the triangles for the resultant velocity questions. Because i know how to get the answer but because my triangle looks different to the mark schemes it's not correct.
Thanks
Original post by afghan-superman
Hi, so MM1B exam for AQA is on Tuesday. And basically, i dont understand how to properly draw the triangles for the resultant velocity questions. Because i know how to get the answer but because my triangle looks different to the mark schemes it's not correct.
Thanks


Can you post a question you've done incorrectly, alongside your working out? So that we can point you in the right direction as far the approach is concerned.
Original post by RDKGames
Can you post a question you've done incorrectly, alongside your working out? So that we can point you in the right direction as far the approach is concerned.


A ship travels through water that is moving due east at a speed of 1.4m/s . The ship travels due north relative to the water at a speed of 7m/s . The resultant velocity of the ship is Vm/s on a bearing theta.

Velocity of the water is 1.4 m/s
Velocity of the ship relative to the water 7m/s
(a)Find V.


(b)Find theta, giving your answer as a three-figure bearing, correct to the nearest degree.




Sorry dont have my working I'm out at the moment. the thing i struggle with mostly is that sometimes they leave theta by itself and other times they subtract it from e.g. 180 degrees. The questions are worded the same but why is this the case?
Thanks for your help :smile:
Original post by afghan-superman
A ship travels through water that is moving due east at a speed of 1.4m/s . The ship travels due north relative to the water at a speed of 7m/s . The resultant velocity of the ship is Vm/s on a bearing theta.

Velocity of the water is 1.4 m/s
Velocity of the ship relative to the water 7m/s
(a)Find V.


(b)Find theta, giving your answer as a three-figure bearing, correct to the nearest degree.




Sorry dont have my working I'm out at the moment. the thing i struggle with mostly is that sometimes they leave theta by itself and other times they subtract it from e.g. 180 degrees. The questions are worded the same but why is this the case?
Thanks for your help :smile:


A good start is to begin sketching what they said. So first we are told that the velocity of water is due east at 1.4m/s so we have:

Spoiler


Then we are told that the velocity of the boat relative to water is 7m/s due north so we have:

Spoiler



Now the resultant velocity is the one that joins the tail of the red vector to the tip of the blue vector:

Spoiler



So to find what the resultant velocity V is, all you have to do is apply Pythagoras' Theorem here.

Then you are asked to find the bearing of V, which is this purple angle because bearings are always measured from the NORTH line CLOCKWISE:

Spoiler



So to work it out, quite simply find the angle between the green vector and the red vector, then take it away from 90. A bearing is a 3 digit number so you need to represent it as such.


Makes sense?? Have a go at a different question with this approach and see if you get it.
Original post by RDKGames
A good start is to begin sketching what they said. So first we are told that the velocity of water is due east at 1.4m/s so we have:

Spoiler


Then we are told that the velocity of the boat relative to water is 7m/s due north so we have:

Spoiler



Now the resultant velocity is the one that joins the tail of the red vector to the tip of the blue vector:

Spoiler



So to find what the resultant velocity V is, all you have to do is apply Pythagoras' Theorem here.

Then you are asked to find the bearing of V, which is this purple angle because bearings are always measured from the NORTH line CLOCKWISE:

Spoiler



So to work it out, quite simply find the angle between the green vector and the red vector, then take it away from 90. A bearing is a 3 digit number so you need to represent it as such.


Makes sense?? Have a go at a different question with this approach and see if you get it.

Thanks alot for the help!
For this question i used the SOH CAH TOA rule and done tan(theta)=7/1.4 and i got the answer without subtracting it from 90.
Original post by afghan-superman
Thanks alot for the help!
For this question i used the SOH CAH TOA rule and done tan(theta)=7/1.4 and i got the answer without subtracting it from 90.

Actually dont worry lol i id subtract is from 90 sorry.
Thanks so much for the help thoigh
Original post by afghan-superman
Thanks alot for the help!
For this question i used the SOH CAH TOA rule and done tan(theta)=7/1.4 and i got the answer without subtracting it from 90.


That's not the correct answer though as then your angle is 78.69 which is not the bearing of the resultant velocity. You need to subtract this answer from 90.

Your angle that you find is α\alpha here, but you want θ\theta

Spoiler

(edited 6 years ago)

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