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Homework Help: Confused on next step in finding instantaneous velocity

  1. Oct 4, 2009 #1
    1. The problem statement, all variables and given/known data
    In a time of t seconds, a particle moves a distance of s meters from its starting point, where s=3t[tex]^2[/tex]+13.

    Find the average velocity between t=1 and t=1+h if
    (i) h=0.1 (ii) h=0.01 (iii) h=0.001

    Use the obtained answers to estimate the instantaneous velocity of the particle at time t=1.

    The average velocity between t=1 and t=1+h
    (i) when h=0.1 is ____ m/s;
    (ii) when h=0.01 is ____ m/s;
    (iii) when h=0.001 is ____ m/s.

    The instantaneous velocity appears to be ______ m/s

    2. Relevant equations

    3. The attempt at a solution
    h=0.1 3(0.1)[tex]^2[/tex]+13=13.03
    h=0.01 3(0.01)[tex]^2[/tex]+13=13.0003
    h=0.001 3(0.001)[tex]^2[/tex]+13=13

    What am I supposed to do next in this problem?
  2. jcsd
  3. Oct 4, 2009 #2


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    You'd better back up and rethink the first part. The average speed between t=1 and t=1+h is s(1+h)-s(1) divided by the elapsed time (h). It's just distance divided by time.
  4. Oct 4, 2009 #3
    If I do s(1+0.1)-s(1)/0.1 that gives me s(0.1)/0.1

    I don't get it. How does that work?
  5. Oct 4, 2009 #4


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    Nooo. s(1+0.01)-s(1) is NOT s(0.01). s(1+0.01)-s(1)=(3*(1.01)^2+13)-(3*1^2+13). That's not the same as (3*(0.01)^2+13). Check the two expressions using your calculator!
    Last edited: Oct 4, 2009
  6. Oct 4, 2009 #5
    I got s(0.1)=0.63, s(0.01)=0.0603, and s(0.001) was too small a number it appeared 0. I don't get how this gets the instantaneous velocity. Do you multiply by 10 and 100 and 1000.. etc..?
  7. Oct 4, 2009 #6


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    The average speed from t=1 to t=1.1 (h=0.1) is (s(1.1)-s(1))/(0.1)=((3*(1.1)^2+13)-(3*1^2+13))/0.1. That is 6.3 by my reckoning. Can you do the other ones, please? There is no reason to compute s(0.1), and even if there were you wouldn't get 0.63. 1.001 isn't so small that 1.001^2 should round off to 1.
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