Motion in 2 or 3 Dimensions:Year 2 Mechanics Watch

aleenapaul
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Hi,
Can anyone explain to me how I should solve this question?
At time t=T the particle is moving at right angles to its initial direction of motion. Find the value of T and the distance of the particle from its initial position at this time.
I calculated the initial velocity=20i+10j. But I am not sure how to find the right angles to the initial velocity.
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ghostwalker
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(Original post by aleenapaul)
Hi,
Can anyone explain to me how I should solve this question?
At time t=T the particle is moving at right angles to its initial direction of motion. Find the value of T and the distance of the particle from its initial position at this time.
I calculated the initial velocity=20i+10j. But I am not sure how to find the right angles to the initial velocity.
You've clearly only posted part of the question - but we can live with that for now.

Direction of motion is velocity.

So, the initial velocity is at right angles to the velocity at time t=T.

If two vectors are at right angles, then their dot product is zero, and that's what you need to use.
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aleenapaul
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(Original post by ghostwalker)
You've clearly only posted part of the question - but we can live with that for now.

Direction of motion is velocity.

So, the initial velocity is at right angles to the velocity at time t=T.

If two vectors are at right angles, then their dot product is zero, and that's what you need to use.
Thanks for the help but I am not sure what you mean by the dot product.
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ghostwalker
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(Original post by aleenapaul)
Thanks for the help but I am not sure what you mean by the dot product.
OK, so you've not covered the dot product yet.

In that case, an alternative method, for 2D at least. Think of the vectors as being straight line graphs. When two lines intersect at right angles, the product of their gradients is -1. And use that. In this case the gradient of a vector can be thougth of as the y-component divided by the x-component.
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aleenapaul
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(Original post by ghostwalker)
OK, so you've not covered the dot product yet.

In that case, an alternative method, for 2D at least. Think of the vectors as being straight line graphs. When two lines intersect at right angles, the product of their gradients is -1. And use that. In this case the gradient of a vector can be thougth of as the y-component divided by the x-component.
I tried doing it and got t=0.75s but the answer should be 5s
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ghostwalker
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(Original post by aleenapaul)
I tried doing it and got t=0.75s but the answer should be 5s
Can't do anything with that. You need to post the full question AND your working.
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aleenapaul
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(Original post by ghostwalker)
Can't do anything with that. You need to post the full question AND your working.
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Size:  127.7 KB it's question 15
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aleenapaul
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For the gradient of the initial velocity 20i +10j I got 0.5 by dividing 10/20.so the perpendicular gradient has to be -2 so the new velocity has to be 20i-10j.
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RDKGames
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(Original post by aleenapaul)
For the gradient of the initial velocity 20i +10j I got 0.5 by dividing 10/20.so the perpendicular gradient has to be -2 so the new velocity has to be 20i-10j.
But your new velocity has gradient of -0.5 rather than -2.

It should be 20i - 40j.


Important to note the i component stays constant in this problem.
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aleenapaul
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(Original post by RDKGames)
But your new velocity has gradient of -0.5 rather than -2.

It should be 20i - 40j.


Important to note the i component stays constant in this problem
Thanks so much 🙃 I understand it now
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