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Construction Antiderivatives

  1. Nov 25, 2007 #1
    1. The problem statement, all variables and given/known data

    A car starts from rest at time t=0 and accelerates at -0.6t+4 meters/sec^2
    for 0[tex]\leq[/tex]t[tex]\leq[/tex]12. How long does it talke for the car to go 100 meters?

    3. The attempt at a solution
    Since it says the acceleration is -0.6t+4
    I integrated it and ended up with v(t)=-0.3t^2 + 4t + C for the velocity
    but I don't know what to do after this.
    I tried to intergrate it agian so that I can find the distance equation and set that
    equal to 100 but there is an unkown constant C that I have to figure out first.
    HELP!!
     
  2. jcsd
  3. Nov 25, 2007 #2
    okay. Since it syas the car starts from the rest
    v(0)=-0.3(0)^2+4(0)+c=0
    therefore v(t)=-0.3t^2+4t
    I got this far.
    But when I integrate this,
    I get s(t)=-0.1t^3+2t^2+C
    another constant........
    Now what do I do?
     
  4. Nov 25, 2007 #3

    CompuChip

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    Science Advisor
    Homework Helper

    Again, you fill in an initial condition. For example, you can call the position of the car at t = 0 the origin, so s = 0. Then the question, when the car has traveled 100 m, comes down to solving s = 100.
    You can also leave the constant in. Then the question is to solve
    [tex]s(t) - s(0) = 100[/tex]
    for t. Though this is slightly more complicated, you can try it: you'll see that the constant drops out in the end, that is: the requested time is not dependent on the initial point with respect to some special reference point, just on the time difference with respect to the (arbitrary) starting point.

    Basically, it's an arbitrary choice: you have to choose which point you call s = 0, that is, relative to which point you're going to measure your distances... might as well be a logical point. Compare it to this situation: I give you a ruler and ask you to measure the width of your screen. You might put the zero of the ruler on the left side, and read off the length from the right side directly. Or you might just hold the ruler up to the screen, read off the left hand side and right hand side marks, and subtract them. In both cases you will get the correct answer. It's just a matter of which point you call zero, but in the end it's about the difference of the number that is on the right and left side. It just makes life easier -- if you have the choice anyway -- to make the left side be zero (by shifting your ruler, or your function s(t)) such that in the end, you will be subtracting zero instead of some arbitrary number C, to get the difference you want.
     
    Last edited: Nov 25, 2007
  5. Nov 25, 2007 #4
    That means s(0)=0 right? then I got 0 for constant again
    s(t)=-0.1t^3+2t^2
    so I set
    -0.1t^3+2t^2=100

    t^3-20t+1000=0
    now solve for t?
    it seems the answers gonna be pretty complicated,,
    Am I doing it right?
     
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