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Calculus (integration)

  1. Jun 26, 2007 #1
    The velocity of a automatic chuck for moving the wire through a nail making machine may be found from the expression:
    v=0.1t(1-t) where v is velocity (m/s) and t is time (sec)

    i) using integration calculate the distance moved by the chuck between two times when velocity is 0.

    what I did first was change it to 0.1t - 0.1t²

    and when t = 1 you get (0.1*1²/2), which = 0.05 and when t = 0 obviously the answer will be 0
    So I've got the distance moved as 0.05m

    Not really sure if thats right, any help would be great :)

    ii) Verify the results obtained in part (i) by using the approximate numerical method known as the Trapezoidal Rule.

    Not quite as sure with this one

    [​IMG][/URL][/IMG] I think this has something to do with it, I'm also pretty sure I'm meant to use the values 0, 0.2, 0.4, 0.6, 0.8 and 1 with the formula v=0.1t(1-t) those values being t

    As I said I'm a bit unsure so any help would be great :)
  2. jcsd
  3. Jun 26, 2007 #2


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    :confused: How did you get this?? To find the distance moved, you have to integrate [itex] \int v dt [/itex] between the initial and final times.
  4. Jun 30, 2007 #3
    I think he's right....He has put v=dx/dt [After finding values of t from v=0.1t(1-t)....he got t=0,1]
    NOw step by step...
    => dx= [0.1t(1-t)]dt
    INtegrating it You get,
    [x] (limits are 0,x) = 0.1[(t^2 - t^3 /2)] (limits are 0,1)---------------(1)

    In equation (1) its (t raise to power 3)Divided by 2

    Solve it and u get the answer.(x=0.05m)
  5. Jun 30, 2007 #4


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    Integrating t does not give t^2 nor does t^2 give t^3/2. What are you thinking of????
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