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Determine acceleration when velocity is v=x^3-4x^2+7x

  1. Mar 2, 2008 #1
    The velocity of a particle moving along the x axis is given by v=[tex]x^{3}[/tex]-4[tex]x^{2}[/tex]+7x whwere x is in meters and v in m/s. determine the acceleration when x=5m...


    the thing i know is that i just can't differentiate v to get y, since it is a function of x..

    i dont know if chain rule would work..and i dont know how to do it..

    Thanks in advance!
     
  2. jcsd
  3. Mar 2, 2008 #2
    Where are you getting a "y"?

    Just find [tex]\frac{dv}{dt}|_{x=5m}[/tex]
     
    Last edited: Mar 2, 2008
  4. Mar 2, 2008 #3
    sorry..i typed the wrong letter...it should be "a"..

    may you pls explain further.. should i differentiate the equation?
     
  5. Mar 2, 2008 #4
    Well acceleration is defined as a rate of change of velocity: [tex]\frac{d(v)}{dt}[/tex].

    Since acceleration is just a change in velocity, you can differentiate a velocity function to get an acceleration function.

    As to differentiating the function, I don't believe you'd need to use the chain rule at all, because v(t) is not in a form of f(g(x)) where f(x) and g(x) are two functions, it's cases like v(t)=f(g(x)) where the chain rule would be applied in differential calculus.
     
    Last edited: Mar 2, 2008
  6. Mar 2, 2008 #5
    a=dv/dt=3*x^2 - 8*x+7
    a(x=5m)=42 m/s^2
     
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