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Thread: Equation of free fall

  1. #11
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    Quote Originally Posted by prdk21 View Post
    so here is the question:

    "The equations for free fall at the surfaces of Pluto is s = 5.28t^2;

    where s is in meters, and t is in seconds. How long would it

    take an object to fall from rest to reach a velocity of 36 km/h?"

    For some reason this question is really stumping me. I don't know where to start, and I don't even really understand what the
    equations represent. Can someone help me understand this conceptually and
    give me some steps to follow? Thanks!
    When you differentiate the "s" twice - you'll get a constant term.

    That "constant acceleration" makes your problem a special case - where we can use the following equation:

    constant acceleration = (velocityfinal - velocityinital)/(lapsed time)

    continue.....
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  2. #12
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    Quote Originally Posted by prdk21 View Post
    The two things that tripped me up were the units (they aren't equal on each side
    of the equation, so my chemistry background had a panic attack) and the fact
    that the displacement equation was only dependent on time squared.
    5.28 should really be 5.28m/s^2. So in the end, when you multiply (5.28m/s^2)(?s^2) the units will be meters which is fine.

    Suppose you have a constant acceleration, a

    So a=a
    Then v = v0 + atThen S= S0 + .5at2
    I simply integrated to get the next equation.

    It seems as though your S0 = 0 (after all S(0)=0) and your a is 10.56m/s2 . The units really do work out.
    Last edited by Jomo; 01-09-2019 at 09:42 PM.
    A mathematician is a blind man in a dark room looking for a black cat which isn’t there. - Charles R. Darwin

  3. #13
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    Quote Originally Posted by prdk21 View Post
    I guess my biggest confusion is the wording and why the displacement equation
    is only squared when most I have seen have a cubed term.
    Thanks for your response!
    Most of the time in Introductory Physics we use the constant acceleration equations, even if it's a Calculus based class. The displacement equation is given as a series of some power of time times a constant, which have various meanings.
    [tex]s = s_0 + v_0t + \dfrac{1}{2}at^2 + \dfrac{1}{6}jt^3 \text{ ...}[/tex]

    Here, the s stands for displacement, v for velocity, a for acceleration, and j for "jerk." (Yes folks, this really is the name for it.) I've never heard of a problem that uses any more terms than this one. This is possibly the equation you are referring to when you say you've seen the "cubed" term.

    The only other example I've seen is something of the form [tex]s = f(t)[/tex] , where f(t) is some function of time. (Depending on your textbook some other letter than f might be used.) This is the form we usually reserve for Calculus based problems and you can get velocity, acceleration, etc. by taking derivatives.

    -Dan
    It should be possible to explain the laws of Physics to a barmaid. - Albert Einstein

  4. #14
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    Quote Originally Posted by topsquark View Post
    Most of the time in Introductory Physics we use the constant acceleration equations, even if it's a Calculus based class. The displacement equation is given as a series of some power of time times a constant, which have various meanings.
    [tex]s = s_0 + v_0t + \dfrac{1}{2}at^2 + \dfrac{1}{6}jt^3 \text{ ...}[/tex]

    Here, the s stands for displacement, v for velocity, a for acceleration, and j for "jerk." (Yes folks, this really is the name for it.) I've never heard of a problem that uses any more terms than this one. This is possibly the equation you are referring to when you say you've seen the "cubed" term.

    The only other example I've seen is something of the form [tex]s = f(t)[/tex] , where f(t) is some function of time. (Depending on your textbook some other letter than f might be used.) This is the form we usually reserve for Calculus based problems and you can get velocity, acceleration, etc. by taking derivatives.

    -Dan
    Yes, I heard of jerk. It is da/dt if I remember correctly (and according to your formula). Unfortunately my Physics professor never spoke about it at all other than to mention it.

    BTW, I have been meaning to say that it is great to have a Physicist helping out here. THANKS!
    Last edited by Jomo; 01-09-2019 at 10:48 PM.
    A mathematician is a blind man in a dark room looking for a black cat which isn’t there. - Charles R. Darwin

  5. #15
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    Quote Originally Posted by topsquark View Post
    Most of the time in Introductory Physics we use the constant acceleration equations, even if it's a Calculus based class. The displacement equation is given as a series of some power of time times a constant, which have various meanings.
    [tex]s = s_0 + v_0t + \dfrac{1}{2}at^2 + \dfrac{1}{6}jt^3 \text{ ...}[/tex]

    Here, the s stands for displacement, v for velocity, a for acceleration, and j for "jerk." (Yes folks, this really is the name for it.) I've never heard of a problem that uses any more terms than this one. This is possibly the equation you are referring to when you say you've seen the "cubed" term.

    The only other example I've seen is something of the form [tex]s = f(t)[/tex] , where f(t) is some function of time. (Depending on your textbook some other letter than f might be used.) This is the form we usually reserve for Calculus based problems and you can get velocity, acceleration, etc. by taking derivatives.

    -Dan
    Ahh, okay. I must be remembering that from Physics that I took quite a few years ago.
    That was really helpful, thank you so much!

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