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Thread: acceleration/ velocity word problem?

  1. #1
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    acceleration/ velocity word problem?

    A particle moves along the x-axis so that at any time t > 0, its acceleration is given by a(t) = $\displaystyle ln(1+2^t)$. If the velocity of the particle is 2 at time t = 1, then what is the velocity of the particle at time t = 2

    I know what to do but i need help.
    I know I have to take the integral of a(t) = $\displaystyle ln(1+2^t)$

    But i dont know how to evaluate:

    $\displaystyle \int ln(1+2^t) dt$

    integral ln(1+2^x) - Wolfram|Alpha
    wolf fram alpha found the integral but it doesnt do me any good because it comes out all funky
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  2. #2
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    Quote Originally Posted by yoman360 View Post
    A particle moves along the x-axis so that at any time t > 0, its acceleration is given by a(t) = $\displaystyle ln(1+2^t)$. If the velocity of the particle is 2 at time t = 1, then what is the velocity of the particle at time t = 2

    I know what to do but i need help.
    I know I have to take the integral of a(t) = $\displaystyle ln(1+2^t)$

    But i dont know how to evaluate:

    $\displaystyle \int ln(1+2^t) dt$

    integral ln(1+2^x) - Wolfram|Alpha
    wolf fram alpha found the integral but it doesnt do me any good because it comes out all funky
    $\displaystyle
    v(2) = v(1) + \int_1^2 a(t) \, dt
    $

    ... get out your calculator.
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  3. #3
    MHF Contributor chisigma's Avatar
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    The 'hard part' of the work is of course the computation of the integral...

    $\displaystyle \int_{1}^{2} \ln (1+2^{t})\cdot dt$ (1)

    With the substitution of variable $\displaystyle x=1+2^{t} \rightarrow dt= \frac{dx} {\ln 2\cdot (x-1)}$ the integral becomes...

    $\displaystyle \frac{1}{\ln 2}\cdot \int_{3}^{5} \frac{\ln x}{x-1}\cdot dx$ (2)

    ... and with further substitution $\displaystyle \xi= \frac{1}{x} \rightarrow dx= - \frac{ d\xi}{\xi^{2}}$ the integral becomes...

    $\displaystyle - \frac{1}{\ln 2}\cdot \int_{\frac{1}{5}}^{\frac{1}{3}} \frac{\ln \xi}{\xi \cdot (1- \xi)}\cdot d\xi$ (3)


    To be continued in next post...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  4. #4
    MHF Contributor chisigma's Avatar
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    In the previous post we are arrived to the identity...

    $\displaystyle \int_{1}^{2} \ln (1+2^{t})\cdot dt = -\frac{1}{\ln 2}\cdot \int_{\frac{1}{5}}^{\frac{1}{3}} \frac{\ln \xi}{\xi\cdot (1-\xi)}\cdot d\xi$ (1)

    Now for $\displaystyle 0<\xi<1$ is...

    $\displaystyle \frac{\ln \xi}{\xi\cdot (1-\xi)} = \ln \xi \cdot (\frac {1}{\xi} + 1 + \xi + \xi^{2} + \dots )$ (2)

    ... and we can apply the following integration rules...

    $\displaystyle \int \frac{\ln \xi}{\xi}\cdot d\xi= \frac{\ln^{2} \xi}{2} + c$

    $\displaystyle \int \xi^{n}\cdot \ln \xi\cdot d\xi= \xi^{n+1}\{\frac{\ln \xi}{n+1}- \frac{1}{(n+1)^{2}}\} + c$ (3)

    ... and the well known series expansion...

    $\displaystyle \sum_{n=1}^{\infty}\frac{\xi^{n}}{n} = -\ln (1-\xi)$ (4)

    ... to arrive at the result...

    $\displaystyle \int \frac{\ln \xi}{\xi\cdot (1-\xi)}\cdot dx = \frac{\ln^{2} \xi}{2}- \ln \xi \cdot \ln (1-\xi) - \ln \xi \cdot \sum_{n=1}^{\infty} \frac{\xi^{n}}{n^{2}} + c $ (4)

    The series in (4) for $\displaystyle \xi= \frac{1}{3}$ and $\displaystyle \xi=\frac{1}{5}$ has very fast convergence so that using (4) the numerical computation of integral (1) is not to hard ...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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