NaBrO
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#1
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I've got 2s for (a), got 13.5m for (b) - are these correct?
Last edited by NaBrO; 3 months ago
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NaBrO
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I'm trying to boost the thread to get an answer:bawling:
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Hallouminatus
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I think your answer to a) is right. Would you like to share your working for b)?
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NaBrO
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Wait -- do I do:
s = ut + 1/2 at^2
= 14 (2) - 2 (4)
= 20m?
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NaBrO
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Also does the particle move past the point A, get to a further displacement, turn around and get back to A? i.e. speeding up because now velocity is negative and direction is negative?
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Hallouminatus
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(Original post by NaBrO)
Also does the particle move past the point A, get to a further displacement, turn around and get back to A? i.e. speeding up because now velocity is negative and direction is negative?
yes, that's how the particle passes A twice. Part a) asks for the time between those passes, and b) the distance travelled in that time.
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NaBrO
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(Original post by Hallouminatus)
yes, that's how the particle passes A twice. Part a) asks for the time between those passes, and b) the distance travelled in that time.
Thank you so so much :rolleyes:, is my answer to (b) correct?
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Hallouminatus
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(Original post by NaBrO)
Thank you so so much :rolleyes:, is my answer to (b) correct?
Sadly, no. Remember that s is displacement, not distance - what the difference?
Also, u for part b) should be the velocity when P first passes A.
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the bear
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the particle passes thru A at 2.5 and 4.5 seconds. thus the time when it has v = 0 will be at 3.5 seconds.

you can find the distance travelled between 2.5 and 3.5 seconds and then double it to get the answer to b)
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NaBrO
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(Original post by Hallouminatus)
Sadly, no. Remember that s is displacement, not distance - what the difference?
Also, u for part b) should be the velocity when P first passes A.
Got u for part (b) to equal 4. Using what the bear says:
(Original post by the bear)
the particle passes thru A at 2.5 and 4.5 seconds. thus the time when it has v = 0 will be at 3.5 seconds.

you can find the distance travelled between 2.5 and 3.5 seconds and then double it to get the answer to b)
s = ut +1/2 at^2
= 4 (1) + 1/2 (-4) (1)
= 2
Therefore, distance travelled = 2m?

Thank you so much for your help so far :innocent:
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the bear
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(Original post by NaBrO)
Got u for part (b) to equal 4. Using what the bear says:

s = ut +1/2 at^2
= 4 (1) + 1/2 (-4) (1)
= 2
Therefore, distance travelled = 2m?
yes. now double it
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NaBrO
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(Original post by the bear)
yes. now double it
That was such a silly mistake... I knew I had to do that but didn't type it lol. Thank youuuu :bunny2:
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