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Homework Help: Varying Mass:Raindrops

  1. Jun 23, 2016 #1
    1. The problem statement, all variables and given/known data
    xg=x*(dv/dt)+v^2 is a differential equation
    Which has a solution of the form v=at
    Where a is a constant.Find a

    2. Relevant equations


    3. The attempt at a solution
    I assumed 'a' as a function of g
    That is a=kg for some constant k
    And proceeded
    I did get the correct answer a=g/3
    My question is since both a and g are accelerations and constants
    Is it correct to assume one as a function of the other?
    For 2 arbitrary constants a and b, we have (a=kb) for another constant 'k' where a and b are never equal to zero at the same time
    I hope my approach is correct
    Any confirmations and insights are much appreciated!

  2. jcsd
  3. Jun 23, 2016 #2
    I can post the rest of the solution if required
    All I need is some confirmation

  4. Jun 23, 2016 #3


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    Since they are constants, you have not really assumed one is a fixed function of the other. Rather, you have assumed that the ratio between them is also a constant, which is trivially true.
  5. Jun 24, 2016 #4
    Yeah sorry
    That's what I did
    But y=kx (say)
    Doesn't this lead to y=f (x)
    For different values of x, won't we have different values of y
    The only difference with this case and my original one is that
    In my original one, both were constants
    While in this one, both are variables
    X being the independent variable and y being the dependent one

    Is my reasoning correct then??

  6. Jun 24, 2016 #5


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    Say we consider g to be a variable; we might run the same experiment in a different gravitational field. You say you assumed a=kg for some constant k, but you never used the assumption that it was constant. So instead, you could say let a, as function of g, be g times some other function of g, namely k(g). That would be quite valid and general.
    But to solve the problem, you must assume dg/dt=0. You are given da/dt=0, so dk/dt=0. This allows you to substitute k(g)g for a and obtain k=1/3, a constant. So it turns out that k is constant even if g varies.
  7. Jun 27, 2016 #6
    I see what you mean
    Since the first derivative of g is zero,it implies g is a constant
    And we already have da/dt=0 from the problem statement
    So this would force dk/dt=0 meaning k would be a constant
    Thanks a lot for your help

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