Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Calculus word problem

  1. Apr 18, 2006 #1
    "An object moves so that its velocity, v is related to its position. s according to v = (b^2 + 2gs) ^1/2 where b and g are constants. Show that the acceleration of the object is constant."

    I typed out the question exactly as it is. I'm confused because I don't really get what to do. To show that acceleration is constant, I need to get rid of that "s" variable in the question. Acceleration is = to d(velocity)/dt... but from the given function, you can only get d(v) / ds.

    So dv/dt = dv/ds * ds/dt

    But how do I find d(s) / dt? Don't I need a function that has position in terms of time?

    TiA
    Preet
     
  2. jcsd
  3. Apr 18, 2006 #2
    Isn't ds/dt just v?
     
  4. Apr 18, 2006 #3

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yes but v isn't a function of t in this case, it is a function of s
     
  5. Apr 18, 2006 #4
    Does that matter? I mean, a = dv/ds * ds/dt = v dv/ds. If it can be shown that this is constant, then the problem is solved. There is no need to get rid of s.
     
  6. Apr 18, 2006 #5

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Could you please show me how;

    [tex]\frac{dv}{ds} \cdot \frac{ds}{dt} = v\frac{dv}{ds}[/tex]

    Perhaps I'm missing something? It is late after all and I've run out of coffee :grumpy:

    ~H
     
  7. Apr 18, 2006 #6
    dv/ds * ds/dt = ds/dt * dv/ds by commutativity and substituting ds/dt for v gives the result I gave.

    Is there a flaw in my reasoning here?
     
  8. Apr 18, 2006 #7

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Ahhhh :mad: It was so simple I missed it :frown:. My frantic scribblings on paper seem so stupid now.

    [tex]\frac{ds}{dt} = v [/tex]

    You reasoning is perfect e(ho0n3.

    To clarify for the OP;

    You were right by using the chain rule to obtain;

    [tex]\frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt}[/tex]

    What you (and I) didn't spot is that;

    [tex]\frac{ds}{dt} = v \Rightarrow a = \frac{dv}{dt} = v\frac{dv}{ds}[/tex]

    As e(ho0n3 correctly stated.

    ~H
     
    Last edited: Apr 18, 2006
  9. Apr 18, 2006 #8
    Yeah, I feel bad not catching that... it was so obvious =/. Thanks a lot for the help!

    -Preet
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook