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Homework Help: Find velocity, Calculus and rates of change

  1. Oct 17, 2013 #1
    1. The displacement (in meters) of a particle moving in a straight line is given by the equation of motion s=[itex]\frac{1}{t^2}[/itex], where t is measured in seconds. Find the velocity of the particle at times t=a, t=1, t=2, and t=3.

    2. Relevant equations


    3. The attempt at a solution

    The question says the d=[itex]\frac{1}{t^2}[/itex]

    I plug that into V=[itex]\frac{d}{t}[/itex], getting s=[itex]\frac{1}{t^3}[/itex].

    When I input t=a, 1, 2, and 3, I get [itex]\frac{1}{a^3}[/itex], 1, [itex]\frac{1}{8}[/itex], and [itex]\frac{1}{27}[/itex].

    However, the textbook says that the answers are -[itex]\frac{2}{a^3}[/itex], -2, -[itex]\frac{1}{4}[/itex], and -[itex]\frac{2}{27}[/itex].

    Where did it get the -2 from? What am I doing wrong?

    Thanks in advance.
  2. jcsd
  3. Oct 17, 2013 #2


    Staff: Mentor

    The formula you are using is the average velocity. In the problem, you're supposed to use the instantaneous velocity, ##\frac{ds}{dt}##. Your book should have examples of how to find the derivative of the displacement, s.
  4. Oct 17, 2013 #3
    Thanks! According to my textbook, the equation for

    instantaneous rates of change = lim x2→x1 [itex]\frac{f(x2)-f(x1)}{x2-x1}[/itex].

    In my case would that mean the equation would be

    [itex]\frac{Δd}{Δt}[/itex] = lim t2→t1 [itex]\frac{f(d2)-f(d1)}{t2-t1}[/itex] ?

    How does that work into the question?
    Last edited: Oct 17, 2013
  5. Oct 18, 2013 #4


    Staff: Mentor

    $$v = ds/dt = \lim_{h \to 0}\frac{s(t + h) - s(t)}{h}$$

    This formula is equivalent to the one in your book, with x2 = t + h and x1 = t.

    Your function is s = s(t) = 1/t2.

    So s(t + h) = ?
    And s(t) = ?
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