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Math Help - Prove that acceleration is constant

  1. #1
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    Prove that acceleration is constant

    Hello,

    Finding my self very frustrated while working on my calculus homework... seeing if someone can help me through where i am getting stuck!

    Question :

    An object moves so that its velocity, v, is related to its position, s, according to the function v=sqrt(b^2+2gs) where b and g are constants. Show that the acceleration (with respect to time) of the object is constant.

    Thank you!, i would really appreciate it, its the only question im stuck on.
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    Re: Prove that acceleration is constant

    Quote Originally Posted by elittlewood View Post
    Hello,

    Finding my self very frustrated while working on my calculus homework... seeing if someone can help me through where i am getting stuck!

    Question :

    An object moves so that its velocity, v, is related to its position, s, according to the function v=sqrt(b^2+2gs) where b and g are constants. Show that the acceleration (with respect to time) of the object is constant.

    Thank you!, i would really appreciate it, its the only question im stuck on.
    There are two ways. The first is to take the time derivative of v (which is a. Note that ds/dt = v.) Then you have the equation:
    \frac{dv}{dt} = \frac{1}{2} \cdot \frac{1}{\sqrt{b^2 + 2gs}} \cdot 2gv

    How can you simplify this? (Hint: Look at the square root. What is this equal to?)

    Or you could look at the Physics equation:
    v^2 = v_0^2 + 2as
    where v_0 = b and g = a...

    -Dan
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    Re: Prove that acceleration is constant

    Hello thanks for the quick reply !

    Okay, i see what you are doing, but let me clarify so i don't get confused. You have used the chain rule to differentiate v. Now, when you say simply this is where i get a little confused, you have pointed out the sqrt... it is equal to velocity so can you sub the v in ? Would that be the next step?
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    Forum Admin topsquark's Avatar
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    Re: Prove that acceleration is constant

    Quote Originally Posted by elittlewood View Post
    Hello thanks for the quick reply !

    Okay, i see what you are doing, but let me clarify so i don't get confused. You have used the chain rule to differentiate v. Now, when you say simply this is where i get a little confused, you have pointed out the sqrt... it is equal to velocity so can you sub the v in ? Would that be the next step?
    You got it! Then do some simple simplifications.

    -Dan
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    Re: Prove that acceleration is constant

    Thank you! i simplified and was left with g...
    May i ask when you first differentiated you had at the end 2gv . Where did the v come from ?
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    Forum Admin topsquark's Avatar
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    Re: Prove that acceleration is constant

    Quote Originally Posted by elittlewood View Post
    Thank you! i simplified and was left with g...
    May i ask when you first differentiated you had at the end 2gv . Where did the v come from ?
    Sorry.
    \frac{dv}{dt} = \frac{1}{2} \cdot \frac{1}{\sqrt{b^2 + 2gs}} \cdot \frac{d}{dt}(b^2 + 2gs) = \frac{1}{2} \cdot \frac{1}{\sqrt{b^2 + 2gx}} \cdot \frac{d}{dt}(2gs)

    And of course ds/dt = v.

    -Dan
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  7. #7
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    Re: Prove that acceleration is constant

    V = √(b^2+2gs) = ds/dt [ Velocity ]
    (d^2 s)/(dt^2 )= 1/(2 √(b^2+2gs) ) 2g (ds )/dt = 1/(2 √(b^2+2gs) ) 2g √(b^2+2gs) = g that is constant and hence the acceleration is constant
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