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Kinematics: Interesting problem (velocity as a function of distance)

  1. Mar 29, 2009 #1
    I have spent a long time stuggling with an interesting problem which was set up by my brother some time ago:

    Consider an object that starts moving along the line according to the equation below:
    [tex]v(s)=1+{s}^{2}[/tex]
    where s is the total distance travelled by the object and v is the velocity of the object as a function of the distance travelled. respectively.
    Notice that the object starts moving at velocity 1 m/s and the velocity continuously increases as it travels.

    The question is, what is the velocity of the object at some time t?
    [tex]v(t)=?[/tex]
    What is the distance travelled at some time t?
    [tex]s(t)=?[/tex]

    I would greatly appreciate if someone posted a detailed solution to the above problem.
     
  2. jcsd
  3. Mar 29, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi leden! Welcome to PF! :smile:
    erm :redface: … that's not the way this forum works! :rofl:

    Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
     
  4. Mar 29, 2009 #3

    Doc Al

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    Hint: v = ds/dt. Set up the integral and solve for s(t) first.
     
  5. Mar 29, 2009 #4
    Re: Welcome to PF!

    I am aware of that..
    But the problem is that am not so much familiar with calculus and when it comes to physics I don't know what to do when integrating/differentiating some functions which is dependent on many variables...

    The first things that bothers me is whether it is correct that
    [tex]v(s)=ds/dt[/tex].
    Ok, have tried the following...
    [tex]v(s)=\frac{ds}{dt} = 1 + {s}^{2}[/tex]
    [tex]ds=dt+{s}^{2}dt[/tex]
    [tex]\int ds=\int dt + \int{s}^{2}dt[/tex]
    [tex]s=t+\int{s}^{2}dt[/tex]
    I don't know what to do know, beacause I'm going into cycles...
     
    Last edited: Mar 29, 2009
  6. Mar 29, 2009 #5

    Doc Al

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    Re: Welcome to PF!

    OK.
    Nah... do this:

    [tex]\frac{ds}{1 + {s}^{2}} = dt[/tex]

    (Now you can integrate.)
     
  7. Mar 29, 2009 #6
    Thanks,
    let me try to solve now...
    [tex]\frac{ds}{1+{s}^2}=dt[/tex]
    [tex]\int\frac{ds}{1+{s}^2}=\int dt[/tex]

    [tex]\arctan s=t[/tex]

    [tex]s(t)=\tan t[/tex]

    Now
    [tex]v(t)=\frac{ds}{dt}=\frac{\tan t}{dt}=\frac{1}{{\cos}^{2}t}[/tex]

    This is strange, because it seems that velocity is practically infinity at point t = pi/2.
     
    Last edited: Mar 29, 2009
  8. Mar 30, 2009 #7

    tiny-tim

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    So is s … what's strange about that? :smile:
     
  9. Mar 30, 2009 #8
    OK but how is it possible that
    [tex]

    v(s)=\frac{ds}{dt}
    [/tex]
    and
    [tex]
    v(t)=\frac{ds}{dt}
    [/tex]

    Velocity, by the definition, is the first derivative of position, but as the function of time.
    So you can't say that velocity as the function of position is also the first derivative of position.
    Correct me if I'm wrong, but that doesn't make any sense to me :)
     
  10. Mar 30, 2009 #9

    Doc Al

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    I'm not seeing the problem. v ≡ ds/dt, which can be expressed as a function of time or of distance. So?
     
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