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# M1 question - slightly confused watch

1. The question is from the MEI paper from June 2008 and asks you to "Calculate the distance traveled by the particle from t = 0 to t = 4."

after giving you the equation for x: x = 10+36t+3t^2-2t^3.

i took the value of x at t=0 from the value of x at t=4, yet in the answer sheet its says (because the time limit is stated as −4 ≤ t ≤ 6 for t in the equation) that i should have used these values instead. here is the link to the paper to check it, i may have read it wrong http://www.mei.org.uk/files/papers/m108ju_vik3.pdf

but surely if it only asks for the "distance" between t=o and t=4 then the answer of 64m i got should be correct??
2. What you've calculated is the distance between the start and end points. This isn't the same as the distance the particle's travelled; I can travel all the way around the world and end up back where I've started, so by your method I've travelled 0m. The reason is that the distance function goes negative at x=3, so you need to split it up into the bit with 0<x<3 and 3<x<4.
3. (Original post by SebastianTindall)
The question is from the MEI paper from June 2008 and asks you to "Calculate the distance traveled by the particle from t = 0 to t = 4."

after giving you the equation for x: x = 10+36t+3t^2-2t^3.

i took the value of x at t=0 from the value of x at t=4, yet in the answer sheet its says (because the time limit is stated as −4 ≤ t ≤ 6 for t in the equation) that i should have used these values instead. here is the link to the paper to check it, i may have read it wrong http://www.mei.org.uk/files/papers/m108ju_vik3.pdf

but surely if it only asks for the "distance" between t=o and t=4 then the answer of 64m i got should be correct??
Try drawing the graph for velocity. HINT: Distance is the integral of velocity where velocity = +ve.
4. (Original post by Noble.)
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(Original post by SamGrainger95)
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Come on guys, ignore the troll and focus on the question in hand (unless you've cleared it up now)

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