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Thread: 1st & 2nd Derivative Physics word problem

  1. #1
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    1st & 2nd Derivative Physics word problem

    I'm stuck on this problem. Can someone help me with where to start?

    A body moves in such a way that the space described in the time t is given by $\displaystyle s=t^n$, where n is a constant. Find the value of n when the velocity is doubled from the 5th to the 10th second; find it also when the velocity is numerically equal to the acceleration at the end of the 10th second.
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  2. #2
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    Hello, dbakeg00!

    You're expected to know that: .$\displaystyle \begin{array}{ccc}v &=& \dfrac{ds}{dt} \\ \\[-3mm] a &=& \dfrac{dv}{dt} \end{array}$


    A body moves in such a way that its position at time $\displaystyle t$ is given by: .$\displaystyle s\:=\:t^n$, where n is a constant.

    (a) Find $\displaystyle n$ when the velocity is doubled from the 5th to the 10th second.
    We have: .$\displaystyle v(t) \:=\:n\,t^{n-1}$

    Then: .$\displaystyle \begin{Bmatrix}v(5) &=& n\,5^{n-1} \\ v(10) &=& n\,10^{n-1} \end{Bmatrix}$


    $\displaystyle v(10)$ is twice $\displaystyle v(5)\!:\;\;n\,10^{n-1} \:=\:2\cdot n\,5^{n-1} \quad\Rightarrow\quad \frac{10^{n-1}}{5^{n-1}} \:=\:2 \quad\Rightarrow\quad 2^{n-1} \:=\:2^1$

    Therefore: .$\displaystyle n-1 \:=\:1 \quad\Rightarrow\quad n \:=\:2$




    (b) Find $\displaystyle n$ when the velocity is numerically equal to the acceleration at the end of the 10th second.
    We have: .$\displaystyle \begin{Bmatrix}v(t) &=& n\,t^{n-1} \\ a(t) &=& n(n-1)t^{n-2} \end{Bmatrix}$

    Then: .$\displaystyle \begin{Bmatrix} v(10 &=& n\,10^{n-1} \\ a(10) &=& n(n-1)\,10^{n-2} \end{Bmatrix}$


    These two quantities are equal: .$\displaystyle n(n-1)\,10^{n-2}\;=\;n\,10^{n-1}$


    Divide by $\displaystyle n\,10^{n-2}\!:\;\;n-1 \:=\:10$


    . . . . . Therefore: .$\displaystyle n \:=\: 11$

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  3. #3
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    Thanks for the quick reply and for your answer, it helped a lot. One more thing...


    These two quantities are equal: .$\displaystyle n(n-1)\,10^{n-2}\;=\;n\,10^{n-1}$


    Divide by $\displaystyle n\,10^{n-2}\!:\;\;n-1 \:=\:10$


    . . . . . Therefore: .$\displaystyle n \:=\: 11$
    So when you divide $\displaystyle n\,10^{n-1}$ by $\displaystyle n\,10^{n-2}$, how do you come up with 10 ?(sorry, I'm a little rusty on division by exponents..I know you should subtract them, but I am having a difficult time seeing this one)
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  4. #4
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    Nevermind...I'm an idiot! I see it now. Thanks again for the help.
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