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    How do you do this Q:

    A particle moves in a straight line with velocity v given by v=3t-4t^3+1 at time t s after passing throught a fixed point O in that line.

    I know that the velocity is 0 when t=1

    and the accerelation is -45 when t=2

    Find

    c) maximum velocity ANS=2

    d) its distance from O when t=2 ANS=-8

    On C, don't you differientate v to get a, make it equal to 0, then solve, you should get 2 but I get a 1/2

    Don't know what to do a d)
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    (c) asks for maximum velocity. You've found what t is, now work out v
    (d) you need to integrate the velocity to find the displacement
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    (Original post by emmaxoxo)
    How do you do this Q:

    A particle moves in a straight line with velocity v given by v=3t-4t^3+1 at time t s after passing throught a fixed point O in that line.

    I know that the velocity is 0 when t=1

    and the accerelation is -45 when t=2

    Find

    c) maximum velocity ANS=2

    d) its distance from O when t=2 ANS=-8

    On C, don't you differientate v to get a, make it equal to 0, then solve, you should get 2 but I get a 1/2

    Don't know what to do a d)
    for c), yes you do get a half, but this is the time. So just sub it back into the velocity formula to get the velocity.

    For D, integrating V gives X (when dealing with t)
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    oh right thanks got it
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    (Original post by ian.slater)
    (c) asks for maximum velocity. You've found what t is, now work out v
    (d) you need to integrate the velocity to find the displacement
    Ok, New Question

    Particle P moves in a striaght line such that t s its accerleation is given by a=2t-2, it passes through a point O on the line t=0 with v=1

    find the distance of P from O when it's at rest.

    Do you find v, integrate it to get x, then sub t=0, to find x at rest??? I get the answer 0 whereas it should be a 1/3
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    (Original post by emmaxoxo)
    Ok, New Question

    Particle P moves in a striaght line such that t s its accerleation is given by a=2t-2, it passes through a point O on the line t=0 with v=1

    find the distance of P from O when it's at rest.

    Do you find v, integrate it to get x, then sub t=0, to find x at rest??? I get the answer 0 whereas it should be a 1/3
    No, you integrate a to find V, then you equate that to 0, since you want V to be 0. Ie when it is at rest v=0. This will give u the value of t.

    Integrate V again to give X formula, then sub the value of t into the x formula to get the Distance
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    (Original post by 2710)
    No, you integrate a to find V, then you equate that to 0, since you want V to be 0. Ie when it is at rest v=0. This will give u the value of t.

    Integrate V again to give X formula, then sub the value of t into the x formula to get the Distance
    OP - Don't forget when you integrate a to get v, that you need to find out what the constant of integration is. You're told that v=1 when t=0 so use that.
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    (Original post by 2710)
    No, you integrate a to find V, then you equate that to 0, since you want V to be 0. Ie when it is at rest v=0. This will give u the value of t.

    Integrate V again to give X formula, then sub the value of t into the x formula to get the Distance
    Thanks. Got it.

    Last Question.

    A particle moves along the x axis with velocity at time t given by v=5t^2+2t

    find the distance it moves in the 2nd second (i.e. t=2)

    Don't you just intergrate v to find x. But how do you find out what C is on x, what are the restraints (i.e. x=0,t=0), therefore c=0, then just sub t=2 into x. I get 17 and 1/3, whereas it should be 14 2/3?
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    (Original post by emmaxoxo)
    Thanks. Got it.

    Last Question.

    A particle moves along the x axis with velocity at time t given by v=5t^2+2t

    find the distance it moves in the 2nd second (i.e. t=2)

    Don't you just intergrate v to find x. But how do you find out what C is on x, what are the restraints (i.e. x=0,t=0), therefore c=0, then just sub t=2 into x. I get 17 and 1/3, whereas it should be 14 2/3?
    So you have to finds the distance between times 2-3 s. So x=0 when t=2. Find c with that, and try it again. I think this is how you do it anyways.
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    Distance moved in 2nd second is x(2) - x(1)

    Fortunately, whatever constant of integration you use will disappear when you work this out. In other words you can use whatever original you want.

    If I were doing this question, I'd be nit-picky enough to check that the particle doesn't change direction between t=1 and t=2, but I don't expect M1 is that awkward!
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    (Original post by 2710)
    So you have to finds the distance between times 2-3 s. So x=0 when t=2. Find c with that, and try it again. I think this is how you do it anyways.
    what so find the distance it moves in the 2nd second does not mean t=2, it means the distance between second 2 and 3 ???

    So C= - 17 and 1/3

    then you sub in t=3 in X and t=2 in X again, and subtract them? if so i'm still getting the wrong ANS
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    (Original post by ian.slater)
    Distance moved in 2nd second is x(2) - x(1)

    Fortunately, whatever constant of integration you use will disappear when you work this out. In other words you can use whatever original you want.

    If I were doing this question, I'd be nit-picky enough to check that the particle doesn't change direction between t=1 and t=2, but I don't expect M1 is that awkward!
    Sorry I keep getting forgetting this is M2.

    Ok I have got the amswer right by saying the C=0, this true then? But why does C=0 ??
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    Woops sorry, yeh, ian is right. Between the 1 and 2nd second
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    (Original post by emmaxoxo)
    Sorry I keep getting forgetting this is M2.

    Ok I have got the amswer right by saying the C=0, this true then? But why does C=0 ??
    Because the question doesn't tell you where the particle is at t=0, or enough information to work it out, you can be sure that they will not ask you anything that needs you to know that.

    So you can say 'let x=0 when t=0' i.e. set your own origin for measurement.
 
 
 
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