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I may have forgotten how to do vector over the 2 week break.......

Particle at (3i-7J) travels 50ms^-1 parallel to (7i-24j)

calculate velocity
position vector after 2 seconds
position vector after t seconds and distance from the origin after 10 seconds

@Zacken help pls i'm going to die from too much homework
Original post by thefatone
I may have forgotten how to do vector over the 2 week break.......

Particle at (3i-7J) travels 50ms^-1 parallel to (7i-24j)

calculate velocity
position vector after 2 seconds
position vector after t seconds and distance from the origin after 10 seconds

@Zacken help pls i'm going to die from too much homework



What does it mean by parallel to that point?

Can you post the actual question, like a picture?
(edited 8 years ago)
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thefatone
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Original post by Dinasaurus
What does it mean by parallel to that point?


no clue, all help is useful
Original post by thefatone
no clue, all help is useful


Have you got a picture of the question?
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thefatone
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Original post by Dinasaurus
What does it mean by parallel to that point?

Can you post the actual question, like a picture?


everyone in the class wrote down the question and that's all we were given... no sheet given, he just dictated and we wrote it down :/
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B_9710
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Original post by thefatone
I may have forgotten how to do vector over the 2 week break.......

Particle at (3i-7J) travels 50ms^-1 parallel to (7i-24j)

calculate velocity
position vector after 2 seconds
position vector after t seconds and distance from the origin after 10 seconds

@Zacken help pls i'm going to die from too much homework


The vector 7i−24j 7 \mathbf{i}-24 \mathbf{j} has magnitude 25, but the particle is travelling at 50ms-1 so every second the particle moves a displacement of 14i−48j 14\mathbf{i}-48\mathbf{j} .

So the particle starts at 3i−7j 3\mathbf{i}-7\mathbf{j} and after t seconds it has moved 14ti−48tj 14t\mathbf{i}-48t\mathbf{j} so we can represent the path of the particle as
r=(3+14t)i−(7+48t)j r=(3+14t)\mathbf{i}-(7+48t)\mathbf{j} , where t t is the time the ball has been travelling in seconds.
(edited 8 years ago)
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thefatone
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Original post by B_9710
The vector 7i−24j 7 \mathbf{i}-24 \mathbf{j} has magnitude 25, but the particle is travelling at 50ms-1 so every second the particle moves a displacement of 14i−48j 14\mathbf{i}-48\mathbf{j} .


so i just double it?
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B_9710
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Original post by thefatone
so i just double it?


For the velocity yes. They give you the speed and direction and from that you can always find velocity as I have already shown.
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KaylaB
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Original post by thefatone
so i just double it?


From all that I can remember at the moment, a vector is parallel to another vector if it is a multiple of it
Like 4i-6j is parallel to 40i-60j as 40i-60j = 10(4i-6j)
Reply 9
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Zacken
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Original post by thefatone
...


Think of it like this, it moves 50 metres per second in the direction of the unit vector 1∣7i−24j∣(7i−24j)\frac{1}{|7i - 24j|} (7i - 24j) so its velocity is 5025(7i−24j)\frac{50}{25}(7i - 24j) and hence its position can be written the way B_9710 wrote it.

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