# Maths - Mechanics

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#1
A particle, P, moves along a straight line. At time t seconds t>= 0, the displacement , x m, of P from the origin O is given by x =t^3 - 11t^2+24t

Find the greatest distance that P is from the origin in the first 5 seconds of motion.

So I have integrated ( sorry meant differentiated) it to find time when v=0 and I got t=4/3 and t=0

I now have three times but I dont know what to do next.
Do i add all the distances of the three times or pick the time with the greatest distance.

Any help will be appreciated
Last edited by 2021alevels2021; 1 month ago
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1 month ago
#2
(Original post by 2021alevels2021)
A particle, P, moves along a straight line. At time t seconds t>= 0, the displacement , x m, of P from the origin O is given by x =t^3 - 11t^2+24t

Find the greatest distance that P is from the origin in the first 5 seconds of motion.

So I have integrated it to find time when v=0 and I got t=4/3 and t=0

I now have three times but I dont know what to do next.
Do i add all the distances of the three times or pick the time with the greatest distance.

Any help will be appreciated
You have been given an expression for displacement. Velocity is the rate of change of displacement, so to find velocity you need to differentiate (not integrate) the given expression.
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#3
(Original post by old_engineer)
You have been given an expression for displacement. Velocity is the rate of change of displacement, so to find velocity you need to differentiate (not integrate) the given expression.
yeah sorry I meant I differentiated. What do i do next to work out the maximum distance
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1 month ago
#4
(Original post by 2021alevels2021)
yeah sorry I meant I differentiated. What do i do next to work out the maximum distance
You said in your first post that you had three time when v = 0, then you mentioned two of them. There should actually be two times, and one of them isn't t = 0. Perhaps have another look at your working, and check by plugging your derived times into your expression for v to make sure it is really zero when you think it is. Post your working if stuck.
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#5
(Original post by old_engineer)
You said in your first post that you had three time when v = 0, then you mentioned two of them. There should actually be two times, and one of them isn't t = 0. Perhaps have another look at your working, and check by plugging your derived times into your expression for v to make sure it is really zero when you think it is. Post your working if stuck.
I redid it and got t = 6 and t=4/3
but since the range of time is from one to five I ignored 6.

I also used the extremities of the range so when t= 0 and t = 5 . Do i just sub in these values into the original equation and find out which t gives the greatest value?
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1 month ago
#6
(Original post by 2021alevels2021)
I redid it and got t = 6 and t=4/3
but since the range of time is from one to five I ignored 6.

I also used the extremities of the range so when t= 0 and t = 5 . Do i just sub in these values into the original equation and find out which t gives the greatest value?
Yep.

Note: You're looking for the greatest distance (an unsigned quantity), and x is displacement (a signed quantity). So, you want the greatest absolute value of x.
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