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Math Help - This confused me. Please help

  1. #1
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    This confused me. Please help

    I got an answer but I am not entirely sure if i was right. can someone please help to verify my answers. thanks

    For a body falling under gravity in air, the rate of change of velocity is given by dV/dt = -k(V-P) where P and K are constants. This has a solution: V=P+Ae^(-kt). If the rate of change by dV/dt = -0.02(V-490), find:

    a) an equation for V, remembering that initially V=0

    b) the velocity after 10 seconds (2dp)

    c) the maximum velocity which could be achieved, assuming no outside interference.
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by chrisgo View Post
    I got an answer but I am not entirely sure if i was right. can someone please help to verify my answers. thanks

    For a body falling under gravity in air, the rate of change of velocity is given by dV/dt = -k(V-P) where P and K are constants. This has a solution: V=P+Ae^(-kt). If the rate of change by dV/dt = -0.02(V-490), find:

    a) an equation for V, remembering that initially V=0

    b) the velocity after 10 seconds (2dp)

    c) the maximum velocity which could be achieved, assuming no outside interference.
    so, would you mind posting what you did so that we can check it for you?
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  3. #3
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    these are my answers

    a) dV/dt = -k(V-P)
    = -0.02(V-490)
    Therefore: k=0.02 and P=490

    V=P+Ae^(-kt)

    k=0.02, P=490
    V=490+Ae^(-0.02t)

    when t=0, V=0
    0=490+Ae^0
    therefore A=490

    V=490-490e^(-0.02t)
    V=490[1-e^(-0.02t)]

    b) when t=0,
    V=490(1-e^-0.2)
    V=88.82193099

    c)maximum velocity when dV/dt=0
    dV/dt=-0.02(V-490)
    0=-0.02(V-490)
    0=V-490
    therefore V=490m/s

    but from the equation:
    V=490[1-e^(-0.02t)] the velocity cannot be equal to 490m/s as e^(any No.) cannot be equal to zero. So how is it that it can have a maximum velocity of 490m/s. What have i done wrong?
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  4. #4
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    Quote Originally Posted by chrisgo View Post
    these are my answers

    a) dV/dt = -k(V-P)
    = -0.02(V-490)
    Therefore: k=0.02 and P=490

    V=P+Ae^(-kt)

    k=0.02, P=490
    V=490+Ae^(-0.02t)

    when t=0, V=0
    0=490+Ae^0
    therefore A=490

    V=490-490e^(-0.02t)
    V=490[1-e^(-0.02t)]
    This is all correct.

    Quote Originally Posted by chrisgo View Post
    b) when t=0,
    V=490(1-e^-0.2)
    V=88.82193099
    This is correct too but remember that they are asking for the answer to 2 decimal places hence v=88.82ms^{-1}

    Quote Originally Posted by chrisgo View Post
    c)maximum velocity when dV/dt=0
    dV/dt=-0.02(V-490)
    0=-0.02(V-490)
    0=V-490
    therefore V=490m/s

    but from the equation:
    V=490[1-e^(-0.02t)] the velocity cannot be equal to 490m/s as e^(any No.) cannot be equal to zero. So how is it that it can have a maximum velocity of 490m/s. What have i done wrong?
    It's an assumption. As e^t \to \infty, it will be a max. It will never approach 490 or exceed it as it is an asymptote so it is assumed to be a maximum (as it will continue 489.9999.....). See the graph attached, it would appear clearer visually.
    Attached Thumbnails Attached Thumbnails This confused me. Please help-graph.jpg  
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  5. #5
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by chrisgo View Post
    these are my answers

    a) dV/dt = -k(V-P)
    = -0.02(V-490)
    Therefore: k=0.02 and P=490

    V=P+Ae^(-kt)

    k=0.02, P=490
    V=490+Ae^(-0.02t)

    when t=0, V=0
    0=490+Ae^0
    therefore A=490 (take note of that..)

    V=490-490e^(-0.02t) (correct!)
    V=490[1-e^(-0.02t)]

    b) when t=0, (oops, t=10)
    V=490(1-e^-0.2)
    V=88.82193099

    c)maximum velocity when dV/dt=0
    dV/dt=-0.02(V-490)
    0=-0.02(V-490)
    0=V-490
    therefore V=490m/s

    but from the equation:
    V=490[1-e^(-0.02t)] the velocity cannot be equal to 490m/s as e^(any No.) cannot be equal to zero. So how is it that it can have a maximum velocity of 490m/s. What have i done wrong?
    because the object is falling and as it continues to fall, the velocity continues to increase..

    take note of that equation, as t tends to infinity, e^(-0.02t) goes to 0 and so V goes to 490.. this is what the blue statement says..
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