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Definite integrals

  1. Mar 28, 2005 #1
    I don't know how to solve this really easy integral.

    suppose: [tex] a=2 m/s^2 [/tex]

    [tex] a = dv / dt [/tex]

    [tex] \int^b_a a \ dt = \int^b_a dv [/tex]

    [tex] \int^b_a 2 \ dt = v_{b} - v_{a} [/tex]

    [tex] 2t_{b} - 2t_{a} = v_{b} - v_{a} [/tex]

    [tex] v_{b} = v_{a} + 2t_{b} - 2t_{a} [/tex]

    I hope everything is ok up to this point. Then I want to know the position: r, and I need to use the same b and a that I have already use before, but...

    [tex] v = dr / dt [/tex]

    [tex] \int^b_a v \ dt = \int^b_a dr [/tex]

    [tex] \int^b_a ( v_{a} + 2t_{b} - 2t_{a} ) \ dt = r_{b} - r_{a} [/tex]

    [tex] v_{a}t_{b} - v_{a}t_{b} ...... \ ??? = r_{b} - r_{a} [/tex]
     
  2. jcsd
  3. Mar 28, 2005 #2

    James R

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    In your first expression for a, is m the mass and s the displacement? Or is s just a constant?
     
  4. Mar 28, 2005 #3

    StatusX

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    It looks like you're trying to find the velocity and position of an object under accelaeration 2 m/sec^2 at a time tb in terms of its velocity at some earlier time ta. If so, you found vb correctly. But to get the position, you need to integrate the velocity as a function of time, not the velocity at tb. To get v(t), just replace tb with the variable t.
     
  5. Mar 28, 2005 #4
    a = 2 m/s^2

    First part is OK. But when you did the second part you need a general form of v to do the diferentiation. This is how I would do this:

    [tex] d^2 r/dt^2 = a [/tex]

    Integrating once, you have

    [tex] dr/dt = a t + A[/tex] --------------------(1)

    where A is a constant and we can find A using the boundary conditions. You have been given at time t = ta , v = va.

    so, [tex] v_a = a t_a + A => A = v_a - a t_a [/tex]

    so [tex] v(t) = at + ( v_a - a t_a) [/tex]

    Integrate (1) wrt t

    [tex] r = at^2/2 + At + B [/tex] -----------------(2)

    where B is a constant and can be found as earlier using Boundary conditions. at t=ta, r = ra .

    [tex] r_a = at_a^2/2 + At_a + B [/tex]

    Solve for B and subtitute in (2) to get an expression for r(t)
     
  6. Mar 28, 2005 #5

    xanthym

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    .


    Because velocity "v" is a function of time "t", which is indicated by writing it "v(t)", you need to maintain the functional dependence of v(t) into the second set of integrations. Your original work is rewritten below in a manner to do that. Also, equations become more managable if {ta=0}.

    QUESTION: What is the distance "r(t)" traveled from time t=0 to time "t" (for constant acceleration "a")??

    [tex] a \ = \ dv/dt [/tex]

    [tex] \int^{t}_{0} a \ dt \ = \ \int^{v(t)}_{v(0)} \ dv [/tex]

    [tex] at \ = \ v(t) - v(0) [/tex]

    [tex] \color{blue} v(t) \ = \ v(0) + at [/tex]


    [tex] v \ = \ dr/dt [/tex]

    [tex] \int^{t}_{0} v(t) \ dt \ = \ \int^{r(t)}_{r(0)} \ dr [/tex]

    [tex] \int^{t}_{0} \color{blue}\left( v(0) + at \right)\color{black} \ dt \ = \ r(t) - r(0) [/tex]

    [tex] v(0)t + (1/2)at^{2} \ = \ r(t) - r(0) [/tex]

    [tex] \color{red} r(t) \ = \ r(0) + v(0)t + (1/2)at^{2} [/tex]

    Now place {a=(2 m/sec^2)} and values for "r(0)" and "v(0)" into the equation for the final result.


    ~~
     
    Last edited: Mar 29, 2005
  7. Mar 29, 2005 #6
    :confused:. Those are units..
     
  8. Mar 29, 2005 #7

    James R

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    Ah! I completely misread. It's a = 2 metres per second squared. Silly me! I think I just assumed that since the rest of the post was in terms of variables, the acceleration was in terms of variables, too.

    I'd do this similarly to how Gamma did it, above.
     
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