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    Lol. I am so self-hating. Ummm, I have started to teach myself M2, but have come onto a question that is annoying me. No relevant examples in the book, and no way to even work from answer backward, that i can see anyhow...
    Heinemann M2 - ex 1C:

    A particle moves along the x axis with velocity at time t, given by v = 5t^2 + 2t. Find a) the distance travelled in the 2nd second. b) the distance travelled in the 4th second.

    When I tried, I integrated to get an equation in x and t, tried to sub values of 3 and then 2, subtracting one from the other, but I didnt get what they got.

    Someone help, and I will be highly grateful!
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    x(t)
    = distance travelled by time t
    = (int over s from 0 to t) 5s^2 + 2s ds
    = (5/3)t^3 + t^2.

    distance travelled in 2nd second
    = x(2) - x(1)
    = 44/3.

    distance travelled in 4th second
    = x(4) - x(3)
    = 206/3.
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    THANK YOU! I am genuinely stupid. I thought the 2nd second was from t = 2 to t = 3. God. I deserve to be dead or something. Thanks so much for such quick assistance, I will sleep easier now, knowing that it was just my terrible incompetence
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    (Original post by RobbieC)
    Lol. I am so self-hating. Ummm, I have started to teach myself M2, but have come onto a question that is annoying me. No relevant examples in the book, and no way to even work from answer backward, that i can see anyhow...
    Heinemann M2 - ex 1C:

    A particle moves along the x axis with velocity at time t, given by v = 5t^2 + 2t. Find a) the distance travelled in the 2nd second. b) the distance travelled in the 4th second.

    When I tried, I integrated to get an equation in x and t, tried to sub values of 3 and then 2, subtracting one from the other, but I didnt get what they got.

    Someone help, and I will be highly grateful!
    x is distance.

    v = 5t^2 + 2t

    Therefore x = 5/3 t^3 + t^2 + c (where c is constant)

    Assume starts at 0, as find distance traveled since t = 0, therefore x = 0 at t = 0. In which case c = 0.

    So x = 5/3 t^3 + t^2

    and at t = 2 seconds...

    x = 5/3 X 2^3 + 2^2

    x = 5/3 X 8 + 4

    x = 17 and 1/3

    at t = 4 seconds....

    x = 5/3 X 4^3 + 4^2

    x = 5/3 X 64 + 16

    x = 122 and 2/3

    Oh and these are distances traveled by that time, so one has to subtract the time before. E.G. t = 2 - t = 1 and t = 4 - t = 3, and you have the equation above. N.B. doing these subtractions results in the constant cancelling out, (xxxxxx + c) - (xxxxxxxx + c) canceles the c out. (That would be my retarded mistake)

    P.S. I have been drinking so take this with a pinch of salt, what question was it in the heineman, as i have the book to hand.
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    OK integrating between bounds works better.

    Int[v] between 1 and 2 (wrt t)
    and
    Int[v] between 3 and 4 (wrt t)

    Where v is a function of t.
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    Thanks Valen, but a Mr. Johnny W already did the question for me. For some reason though, there never seems to be a constant of integration for distance, unless it is specified, I am not entirely sure why. Oh well, Im moving onto calculus with vectors today, should be sickeningly hard

    Then we have CENTRES OF MASS, which looks harder than it first did
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    (Original post by RobbieC)
    Thanks Valen, but a Mr. Johnny W already did the question for me. For some reason though, there never seems to be a constant of integration for distance, unless it is specified, I am not entirely sure why. Oh well, Im moving onto calculus with vectors today, should be sickeningly hard

    Then we have CENTRES OF MASS, which looks harder than it first did
    Probably because at t=0, the object is at the origin.
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    (Original post by RobbieC)
    Then we have CENTRES OF MASS, which looks harder than it first did
    How come? There's not really that much to it. It's not like you have to use integration to find c of m in M2 at all either.
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    Ive just been told that when many shapes are joined on inclined planes and such, it is hard. Using the formulae on their own seems alright, though.

    To me, everything is difficult until I can do it, and do it well.
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    (Original post by mik1a)
    Probably because at t=0, the object is at the origin.
    If x(0) = 0 and v(t) = sin(t) then

    x(t) = (int) v(t) dt = -cos(t) + c,

    so c = 1. You get the same answer from

    x(t) = (int over s from 0 to t) v(s) ds.
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    (Original post by RobbieC)
    Thanks Valen, but a Mr. Johnny W already did the question for me. For some reason though, there never seems to be a constant of integration for distance, unless it is specified, I am not entirely sure why. Oh well, Im moving onto calculus with vectors today, should be sickeningly hard

    Then we have CENTRES OF MASS, which looks harder than it first did
    I realised that after posting, as i was typing up my post when he replied, so didnt see his post.

    You wouldnt have a constant of integration if you integrate between bounds, such as from t = 0 to t = t to give a general formula for the distance traveled since t = 0, or a exact value as i suggested up there. The problems usually specify if the particle dosent start at the origen, and that would be when you have a constant of integration.

    P.S. Calculus with vectors is easy if you just remember to treat each part (i,j,k) as a seperate equation and integrate/differentiate independantly.
 
 
 
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