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Math Help - Simple Linear Motion Problem

  1. #1
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    Cool Simple Linear Motion Problem

    If I know the initial velocity and the distance traveled before coming to a stop (velocity = 0) and assuming the (negative) acceleration is constant, how can I calculate the acceleration?

    E.G.
    I'm in a buggy going 100 feet/sec and there is a light 33 feet in front of me. How much (constant negative) acceleration do I need to come to a stop after 33 feet?

    I can solve other problems, but when I know only the initial velocity and the distance traveled I can't seem to figure out how to calculate the acceleration.

    -Thanks!
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  2. #2
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    Use

    v^2 = u^2 + 2 a s,

    where v is final velocity (zero in this case), u is initial velocity, a is acceleration (negative here) and s is the distance traveled.
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  3. #3
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    Quote Originally Posted by JamesLeighe View Post
    If I know the initial velocity and the distance traveled before coming to a stop (velocity = 0) and assuming the (negative) acceleration is constant, how can I calculate the acceleration?

    E.G.
    I'm in a buggy going 100 feet/sec and there is a light 33 feet in front of me. How much (constant negative) acceleration do I need to come to a stop after 33 feet?

    I can solve other problems, but when I know only the initial velocity and the distance traveled I can't seem to figure out how to calculate the acceleration.

    -Thanks!
    v^2=u^2+2as

    0^2=(100)^2+2a(33)

    a=\frac{-10000}{66}=-151.51\; ft/s^2
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  4. #4
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    Thanks guys!

    Now I just have to look into how it's derived so I can understand why it works.
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  5. #5
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    I guess you mean you want to know how the equation is derived.

    For constant acceleration, a, we have that

    \frac{dv}{dt} = a,

    but using the chain rule we have that

    \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx}

    so

    v \frac{dv}{dx} =a.

    Integrating this wrt x we have

    \int v \, \mathrm{d} v = a \int \, \mathrm{d}x

    and if we replace v and x with dummy variables v' and x' and use the fact that at x'=0, v'=u and at x'=s, v' = v to add the limits we have

    \int_{u}^{v} v' \, \mathrm{d} v' = a \int_{0}^{s} \, \mathrm{d}x'

    which gives us that

    \frac{1}{2} \left(v^2 - u^2 \right) = a s,

    \Leftrightarrow v^2 = u^2 + 2 a s.

    Hope that helps!
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  6. #6
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    Yeah, that's what I really needed. Thanks Doc.
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