Mechanics M2 - Variable Acceleration Question Watch

123456789012
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I'm uncertain of the answer for a linear, variable acceleration question for M2. The question is: A particle P moves along the x-axis. At time t seconds (where t》0) the velocity of P is (3t^2-12t+5) ms^-1 in the direction of x increasing. When t=0, P is at the origin O. Find the distance travelled by P in the interval 3《t《4. The answer in the back of the book is 48m. My teacher says that the answer is 24m. However, I'm very sure that my answer of 2.26m is correct. Could someone please try this question and tell me what they get. Thanks in advance.
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ghostwalker
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(Original post by 123456789012)
I'm very sure that my answer of 2.26m is correct.
Agreed.

See here
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123456789012
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(Original post by ghostwalker)
Agreed.

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Thanks a lot! Very much appreciated!
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a10
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(Original post by ghostwalker)
Agreed.

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Just read this post and i get it now but i had a question, when you find the distance travelled at 3s and 4s both give the value of -12 arent you then supposed to add them together (-12 + -12) to get the total distance giving you a distance of -24m and then because it goes back again multiply by 2 hence giving you the 48m???

Also another question, why do we have to assume the turning point part, when v=0?

Just wondering xD
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123456789012
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(Original post by a10)
Just read this post and i get it now but i had a question, when you find the distance travelled at 3s and 4s both give the value of -12 arent you then supposed to add them together (-12 + -12) to get the total distance giving you a distance of -24m and then because it goes back again multiply by 2 hence giving you the 48m???

Also another question, why do we have to assume the turning point part, when v=0?

Just wondering xD
Hey there, there are two times where the particle changes direction. These two values are 3.52s and 0.42s. You obtain these values by setting v=0. We do this because at maximum displacement dx/dt=0 (therefore v=0 since v=dx/dt). I think you're confusing yourself with displacement and distance. The displacement at t=3 and t=4 is -12. However, this doesn't mean that the particle is stationary between these times - we know this because we know from earlier that the particle reverses direction between t=3 and t=4 (i.e. at 3.52s). I hope I've explained sufficiently.
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a10
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(Original post by 123456789012)
Hey there, there are two times where the particle changes direction. These two values are 3.52s and 0.42s. You obtain these values by setting v=0. We do this because at maximum displacement dx/dt=0 (therefore v=0 since v=dx/dt). I think you're confusing yourself with displacement and distance. The displacement at t=3 and t=4 is -12. However, this doesn't mean that the particle is stationary between these times - we know this because we know from earlier that the particle reverses direction between t=3 and t=4 (i.e. at 3.52s). I hope I've explained sufficiently.
Ahhh right, i just thought that because it said linear along the x axis so i assumed we could add up the distances? Thanks for clearing that up
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