help with vectors

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Thread starter 4 weeks ago
#1
How do I go around answering this question?

'The component of your velocity directly north is 6 m/s. The component directly west is 9 m/s. What is your resultant velocity?'
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4 weeks ago
#2
If you want you can draw a tip-to-tail diagram, which should give you a right-angled triangle.The triangle should help you to find the magnitude of the resultant velocity using simple geometry. (Resultant just means the vector sum)
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Thread starter 4 weeks ago
#3
(Original post by Dash7)
If you want you can draw a tip-to-tail diagram, which should give you a right-angled triangle.The triangle should help you to find the magnitude of the resultant velocity using simple geometry. (Resultant just means the vector sum)
I've done the right angle triangle, but I just don't know what I'm supposed to do after that.
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4 weeks ago
#4
(Original post by Goosha)
I've done the right angle triangle, but I just don't know what I'm supposed to do after that.
So you have two sides of the right angles triangle -what do you know that helps you work out the third side?
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Thread starter 4 weeks ago
#5
I have the third side as 10.8 (1dp), does that mean my velocity is 10.8? and does it have to have a direction etc?
(Original post by GabiAbi84)
So you have two sides of the right angles triangle -what do you know that helps you work out the third side?
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3 weeks ago
#6
yes that is the size of your resultant velocity. It does have a direction, as it is a vector. The direction is from wherever you started drawing the tail of your first component, to the tip of the second component you drew. You can work out the angle in the triangle, and you can use that to find the bearing of the direction of your resultant velocity.

Just to help you out a bit, the direction of your resultant velocity should be towards the North-West direction because of the directions of the two components you have.
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Thread starter 3 weeks ago
#7
(Original post by Dash7)
yes that is the size of your resultant velocity. It does have a direction, as it is a vector. The direction is from wherever you started drawing the tail of your first component, to the tip of the second component you drew. You can work out the angle in the triangle, and you can use that to find the bearing of the direction of your resultant velocity.

Just to help you out a bit, the direction of your resultant velocity should be towards the North-West direction because of the directions of the two components you have.
Thanks for your help! I'm still unsure how I can find or describe the angle. is it 10.8m/s north-west?
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3 weeks ago
#8
(Original post by Goosha)
Thanks for your help! I'm still unsure how I can find or describe the angle. is it 10.8m/s north-west?
The convention is to give it as a bearing - the angle from North measured clockwise
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Thread starter 3 weeks ago
#9
(Original post by Joinedup)
The convention is to give it as a bearing - the angle from North measured clockwise
Okay, tutor has specified the answer to be with north, east, south, west?
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3 weeks ago
#10
(Original post by Goosha)
Okay, tutor has specified the answer to be with north, east, south, west?
In general terms yes you could describe it as 10.8m/s in a north west direction.
To be more precise you could work out the angle that the vector takes from the north point -depends what level you are at whether that would be required.
Last edited by GabiAbi84; 3 weeks ago
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Thread starter 3 weeks ago
#11
(Original post by GabiAbi84)
In general terms yes you could describe it as 10.8m/s in a north west direction.
To be more precise you could work out the angle that the vector takes from the north point -depends what level you are at whether that would be required.
Okay, then how would i find the resultant force?
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3 weeks ago
#12
Resultant force? There will only be a resultant force if you (the object) are accelerating, as F = ma. So if there are more details in the question that tell you whether your velocity is changing or not, you can work out a (acceleration, which might be 0) to find the resultant force F for a given mass.
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