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A position time function!

  1. Apr 6, 2006 #1
    Hello again and here is another problem I can't seem to figure out.

    Chelsea, a calico cat, runs in a straight line chasing a car down the street. Chelsea weighs 36.5 N. and her position along the road is given by
    x(t)=(3.50m/s)t+(0.425m/s^2)t^2+1.00ms/(t+1.00s)
    Calculate Chelsea's kinitec energy at t=2.50s

    The answe is 57.1 joules

    Okay not im not to sure if I'm supposed to find the first or the second derivative. well I compiled it like so
    x(t)=at+bt^2+c/(t+d)
    x'(t)=a+2bt+(t+d)*F((x)c)-c*F((x)t+d)/(t+d)^2 which comes out to
    x'(t)=3.50m/s+2(0.425m/s^2)(2.50s)+(2.5m/s^2)/(2.5m/s+1.00s)^2 = 5.83m
    (m*N)=Joules, 5.83m*36.5 N = 212.8 J <--way off so i tried for a second deriv.
    x''(t)=2(0.425m/s^2+6.25m/s^2/(2.5s*1.00s)^3= 0.996m 0.996m*36.5N=36.3 J <-- still wrong , Where am I messing up can anyone please help me!?!?!
     
  2. jcsd
  3. Apr 6, 2006 #2

    Curious3141

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    What is the formula for kinetic energy in terms of mass and velocity ?

    Is velocity the first differential or second differential of the position wrt time ?

    The differentiation is wrong. What is [tex]\frac{d}{dx}(\frac{1}{x + c})[/tex] ?
     
  4. Apr 6, 2006 #3
    well I used the quotient rule, at the end because it seemed most logical. all we have to do is calculate the amount of joules whic m*N= J and we find the deriv. of the position time function given But, I'm not so sure.
     
  5. Apr 6, 2006 #4

    Curious3141

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    From the way you wrote the expression out, isn't it of the form :

    [tex]3.5t + 0.425t^2 + \frac{1}{t + 1}[/tex]

    rather than

    [tex]\frac{3.5t + 0.425t^2 +1}{t + 1}[/tex] ?

    The former expression does not need quotient rule to differentiate it. Well you can, but there's a much easier way to do it. But first clarify the question. Besides where do all the F(x) etc come from ?
     
  6. Apr 6, 2006 #5
    [tex]3.5t + 0.425t^2 + \frac{1}{t + 1}[/tex] That is the correct equation.

    Im in the same class, by the F(x)'s thats the position time function we are givin. Now Im confused on how to find Kinetic energy. Do I just determine the first derivative? Will that give me K.E(t)?

    Thanks,
    Andy
     
  7. Apr 6, 2006 #6

    Curious3141

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    The first derivative of the displacement will give the velocity.

    The kinetic energy is given by 0.5*mass*(velocity)^2

    The mass can be found from the weight of the cat (given).

    Can you differentiate [tex]\frac{1}{t + 1}[/tex] wrt t

    by chain rule ? Or you can use quotient rule, but chain rule is easier here.
     
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