Word Problem Involving Velocities, Acceleration, Limits












1












$begingroup$


This is the problem:





I took the integral of a(t) to get v(t) and got $v(t)$ = $ln(t+2)$ $-$ $ln(t+3)$ + C . I plugged in the initial conditions and got C = 0. Then when I took the limit, I know it's a telescoping series so everything cancels out except the $ln(2)$ but that's not an answer choice so I'm confused what I did wrong.






calculus integration







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$endgroup$

















    1












    $begingroup$


    This is the problem:





    I took the integral of a(t) to get v(t) and got $v(t)$ = $ln(t+2)$ $-$ $ln(t+3)$ + C . I plugged in the initial conditions and got C = 0. Then when I took the limit, I know it's a telescoping series so everything cancels out except the $ln(2)$ but that's not an answer choice so I'm confused what I did wrong.






    calculus integration







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      This is the problem:





      I took the integral of a(t) to get v(t) and got $v(t)$ = $ln(t+2)$ $-$ $ln(t+3)$ + C . I plugged in the initial conditions and got C = 0. Then when I took the limit, I know it's a telescoping series so everything cancels out except the $ln(2)$ but that's not an answer choice so I'm confused what I did wrong.






      calculus integration







      share|cite|improve this question











      $endgroup$




      This is the problem:





      I took the integral of a(t) to get v(t) and got $v(t)$ = $ln(t+2)$ $-$ $ln(t+3)$ + C . I plugged in the initial conditions and got C = 0. Then when I took the limit, I know it's a telescoping series so everything cancels out except the $ln(2)$ but that's not an answer choice so I'm confused what I did wrong.






      calculus integration




      calculus integration






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 19 '18 at 4:44









      Andrei

      12.4k21128




      12.4k21128










      asked Dec 19 '18 at 4:40









      krauser126krauser126

      474




      474






















          2 Answers
          2






          active

          oldest

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          1












          $begingroup$

          Here I assume that the initial velocity refers to $v(0) = ln frac{2}{3}$:
          $$v(t) = ln frac{t+2}{t+3} = ln frac{1+frac{2}{t}}{1+frac{3}{t}}stackrel{t to infty}{longrightarrow}ln 1 = 0$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            That makes sense. However, why doesn't using the fact that they're telescopic work in this case?
            $endgroup$
            – krauser126
            Dec 19 '18 at 4:53










          • $begingroup$
            @krauser126 Could you please write down the telescoping sum you are referring to? This might clear up what went wrong.
            $endgroup$
            – trancelocation
            Dec 19 '18 at 5:07










          • $begingroup$
            [ln (2) + ln(3) + ln(4) + ...] - [ln (3) + ln (4) + ln (5) + ...] Since its from t = 0 to t = infinity. So wouldnt all the terms in the negative set cancel with those equivalent in the positive set, leaving + ln(2) as the first term left over.
            $endgroup$
            – krauser126
            Dec 19 '18 at 5:15










          • $begingroup$
            @krauser126 This expression would be adding the velocities at specific points: $v(0) + v(1) + cdots $ and does not represent the limit for $t to infty$
            $endgroup$
            – trancelocation
            Dec 19 '18 at 5:26










          • $begingroup$
            @krauser126 Besides this: what you wrote is an undefined expression: $infty - infty$. If you set $v(k) = ln (k+2) - ln (k+3)$ you would get $$Rightarrow sum_{k=0}^n v(k) = sum_{k=0}^n (ln (k+2) - ln (k+3)) = ln 2 - ln (n+3)$$ Now ask yourself whether this reflects the question to be answered.
            $endgroup$
            – trancelocation
            Dec 19 '18 at 5:34





















          0












          $begingroup$

          You have
          $$
          v(t)=lntfrac23+int_0^tfrac1{u^2+5u+6},du=lntfrac23+lnfrac{t+2}{t+3}+lntfrac32,
          $$
          where the $ln tfrac32$ comes from evaluating the antiderivative at $0$. Then
          $$
          lim_{ttoinfty}v(t)=ln tfrac23+ln tfrac32=0.
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I didn't understand a thing you did. Doesn't it tell us that the velocity is initially ln(2/3) so that we can use that as an initial condition to plug t = 0 and solve for C, which I did already and got C = 0. So why did you do that additional step.
            $endgroup$
            – krauser126
            Dec 19 '18 at 4:52










          • $begingroup$
            Now I see that I misread something. Give me a second.
            $endgroup$
            – Martin Argerami
            Dec 19 '18 at 4:56










          • $begingroup$
            You can do it as I did, or you find that $C=0$. In any case, the limit is zero.
            $endgroup$
            – Martin Argerami
            Dec 19 '18 at 4:59











          Your Answer





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          2 Answers
          2






          active

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Here I assume that the initial velocity refers to $v(0) = ln frac{2}{3}$:
          $$v(t) = ln frac{t+2}{t+3} = ln frac{1+frac{2}{t}}{1+frac{3}{t}}stackrel{t to infty}{longrightarrow}ln 1 = 0$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            That makes sense. However, why doesn't using the fact that they're telescopic work in this case?
            $endgroup$
            – krauser126
            Dec 19 '18 at 4:53










          • $begingroup$
            @krauser126 Could you please write down the telescoping sum you are referring to? This might clear up what went wrong.
            $endgroup$
            – trancelocation
            Dec 19 '18 at 5:07










          • $begingroup$
            [ln (2) + ln(3) + ln(4) + ...] - [ln (3) + ln (4) + ln (5) + ...] Since its from t = 0 to t = infinity. So wouldnt all the terms in the negative set cancel with those equivalent in the positive set, leaving + ln(2) as the first term left over.
            $endgroup$
            – krauser126
            Dec 19 '18 at 5:15










          • $begingroup$
            @krauser126 This expression would be adding the velocities at specific points: $v(0) + v(1) + cdots $ and does not represent the limit for $t to infty$
            $endgroup$
            – trancelocation
            Dec 19 '18 at 5:26










          • $begingroup$
            @krauser126 Besides this: what you wrote is an undefined expression: $infty - infty$. If you set $v(k) = ln (k+2) - ln (k+3)$ you would get $$Rightarrow sum_{k=0}^n v(k) = sum_{k=0}^n (ln (k+2) - ln (k+3)) = ln 2 - ln (n+3)$$ Now ask yourself whether this reflects the question to be answered.
            $endgroup$
            – trancelocation
            Dec 19 '18 at 5:34


















          1












          $begingroup$

          Here I assume that the initial velocity refers to $v(0) = ln frac{2}{3}$:
          $$v(t) = ln frac{t+2}{t+3} = ln frac{1+frac{2}{t}}{1+frac{3}{t}}stackrel{t to infty}{longrightarrow}ln 1 = 0$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            That makes sense. However, why doesn't using the fact that they're telescopic work in this case?
            $endgroup$
            – krauser126
            Dec 19 '18 at 4:53










          • $begingroup$
            @krauser126 Could you please write down the telescoping sum you are referring to? This might clear up what went wrong.
            $endgroup$
            – trancelocation
            Dec 19 '18 at 5:07










          • $begingroup$
            [ln (2) + ln(3) + ln(4) + ...] - [ln (3) + ln (4) + ln (5) + ...] Since its from t = 0 to t = infinity. So wouldnt all the terms in the negative set cancel with those equivalent in the positive set, leaving + ln(2) as the first term left over.
            $endgroup$
            – krauser126
            Dec 19 '18 at 5:15










          • $begingroup$
            @krauser126 This expression would be adding the velocities at specific points: $v(0) + v(1) + cdots $ and does not represent the limit for $t to infty$
            $endgroup$
            – trancelocation
            Dec 19 '18 at 5:26










          • $begingroup$
            @krauser126 Besides this: what you wrote is an undefined expression: $infty - infty$. If you set $v(k) = ln (k+2) - ln (k+3)$ you would get $$Rightarrow sum_{k=0}^n v(k) = sum_{k=0}^n (ln (k+2) - ln (k+3)) = ln 2 - ln (n+3)$$ Now ask yourself whether this reflects the question to be answered.
            $endgroup$
            – trancelocation
            Dec 19 '18 at 5:34
















          1












          1








          1





          $begingroup$

          Here I assume that the initial velocity refers to $v(0) = ln frac{2}{3}$:
          $$v(t) = ln frac{t+2}{t+3} = ln frac{1+frac{2}{t}}{1+frac{3}{t}}stackrel{t to infty}{longrightarrow}ln 1 = 0$$






          share|cite|improve this answer











          $endgroup$



          Here I assume that the initial velocity refers to $v(0) = ln frac{2}{3}$:
          $$v(t) = ln frac{t+2}{t+3} = ln frac{1+frac{2}{t}}{1+frac{3}{t}}stackrel{t to infty}{longrightarrow}ln 1 = 0$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 19 '18 at 4:58

























          answered Dec 19 '18 at 4:47









          trancelocationtrancelocation

          12.3k1826




          12.3k1826












          • $begingroup$
            That makes sense. However, why doesn't using the fact that they're telescopic work in this case?
            $endgroup$
            – krauser126
            Dec 19 '18 at 4:53










          • $begingroup$
            @krauser126 Could you please write down the telescoping sum you are referring to? This might clear up what went wrong.
            $endgroup$
            – trancelocation
            Dec 19 '18 at 5:07










          • $begingroup$
            [ln (2) + ln(3) + ln(4) + ...] - [ln (3) + ln (4) + ln (5) + ...] Since its from t = 0 to t = infinity. So wouldnt all the terms in the negative set cancel with those equivalent in the positive set, leaving + ln(2) as the first term left over.
            $endgroup$
            – krauser126
            Dec 19 '18 at 5:15










          • $begingroup$
            @krauser126 This expression would be adding the velocities at specific points: $v(0) + v(1) + cdots $ and does not represent the limit for $t to infty$
            $endgroup$
            – trancelocation
            Dec 19 '18 at 5:26










          • $begingroup$
            @krauser126 Besides this: what you wrote is an undefined expression: $infty - infty$. If you set $v(k) = ln (k+2) - ln (k+3)$ you would get $$Rightarrow sum_{k=0}^n v(k) = sum_{k=0}^n (ln (k+2) - ln (k+3)) = ln 2 - ln (n+3)$$ Now ask yourself whether this reflects the question to be answered.
            $endgroup$
            – trancelocation
            Dec 19 '18 at 5:34




















          • $begingroup$
            That makes sense. However, why doesn't using the fact that they're telescopic work in this case?
            $endgroup$
            – krauser126
            Dec 19 '18 at 4:53










          • $begingroup$
            @krauser126 Could you please write down the telescoping sum you are referring to? This might clear up what went wrong.
            $endgroup$
            – trancelocation
            Dec 19 '18 at 5:07










          • $begingroup$
            [ln (2) + ln(3) + ln(4) + ...] - [ln (3) + ln (4) + ln (5) + ...] Since its from t = 0 to t = infinity. So wouldnt all the terms in the negative set cancel with those equivalent in the positive set, leaving + ln(2) as the first term left over.
            $endgroup$
            – krauser126
            Dec 19 '18 at 5:15










          • $begingroup$
            @krauser126 This expression would be adding the velocities at specific points: $v(0) + v(1) + cdots $ and does not represent the limit for $t to infty$
            $endgroup$
            – trancelocation
            Dec 19 '18 at 5:26










          • $begingroup$
            @krauser126 Besides this: what you wrote is an undefined expression: $infty - infty$. If you set $v(k) = ln (k+2) - ln (k+3)$ you would get $$Rightarrow sum_{k=0}^n v(k) = sum_{k=0}^n (ln (k+2) - ln (k+3)) = ln 2 - ln (n+3)$$ Now ask yourself whether this reflects the question to be answered.
            $endgroup$
            – trancelocation
            Dec 19 '18 at 5:34


















          $begingroup$
          That makes sense. However, why doesn't using the fact that they're telescopic work in this case?
          $endgroup$
          – krauser126
          Dec 19 '18 at 4:53




          $begingroup$
          That makes sense. However, why doesn't using the fact that they're telescopic work in this case?
          $endgroup$
          – krauser126
          Dec 19 '18 at 4:53












          $begingroup$
          @krauser126 Could you please write down the telescoping sum you are referring to? This might clear up what went wrong.
          $endgroup$
          – trancelocation
          Dec 19 '18 at 5:07




          $begingroup$
          @krauser126 Could you please write down the telescoping sum you are referring to? This might clear up what went wrong.
          $endgroup$
          – trancelocation
          Dec 19 '18 at 5:07












          $begingroup$
          [ln (2) + ln(3) + ln(4) + ...] - [ln (3) + ln (4) + ln (5) + ...] Since its from t = 0 to t = infinity. So wouldnt all the terms in the negative set cancel with those equivalent in the positive set, leaving + ln(2) as the first term left over.
          $endgroup$
          – krauser126
          Dec 19 '18 at 5:15




          $begingroup$
          [ln (2) + ln(3) + ln(4) + ...] - [ln (3) + ln (4) + ln (5) + ...] Since its from t = 0 to t = infinity. So wouldnt all the terms in the negative set cancel with those equivalent in the positive set, leaving + ln(2) as the first term left over.
          $endgroup$
          – krauser126
          Dec 19 '18 at 5:15












          $begingroup$
          @krauser126 This expression would be adding the velocities at specific points: $v(0) + v(1) + cdots $ and does not represent the limit for $t to infty$
          $endgroup$
          – trancelocation
          Dec 19 '18 at 5:26




          $begingroup$
          @krauser126 This expression would be adding the velocities at specific points: $v(0) + v(1) + cdots $ and does not represent the limit for $t to infty$
          $endgroup$
          – trancelocation
          Dec 19 '18 at 5:26












          $begingroup$
          @krauser126 Besides this: what you wrote is an undefined expression: $infty - infty$. If you set $v(k) = ln (k+2) - ln (k+3)$ you would get $$Rightarrow sum_{k=0}^n v(k) = sum_{k=0}^n (ln (k+2) - ln (k+3)) = ln 2 - ln (n+3)$$ Now ask yourself whether this reflects the question to be answered.
          $endgroup$
          – trancelocation
          Dec 19 '18 at 5:34






          $begingroup$
          @krauser126 Besides this: what you wrote is an undefined expression: $infty - infty$. If you set $v(k) = ln (k+2) - ln (k+3)$ you would get $$Rightarrow sum_{k=0}^n v(k) = sum_{k=0}^n (ln (k+2) - ln (k+3)) = ln 2 - ln (n+3)$$ Now ask yourself whether this reflects the question to be answered.
          $endgroup$
          – trancelocation
          Dec 19 '18 at 5:34













          0












          $begingroup$

          You have
          $$
          v(t)=lntfrac23+int_0^tfrac1{u^2+5u+6},du=lntfrac23+lnfrac{t+2}{t+3}+lntfrac32,
          $$
          where the $ln tfrac32$ comes from evaluating the antiderivative at $0$. Then
          $$
          lim_{ttoinfty}v(t)=ln tfrac23+ln tfrac32=0.
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I didn't understand a thing you did. Doesn't it tell us that the velocity is initially ln(2/3) so that we can use that as an initial condition to plug t = 0 and solve for C, which I did already and got C = 0. So why did you do that additional step.
            $endgroup$
            – krauser126
            Dec 19 '18 at 4:52










          • $begingroup$
            Now I see that I misread something. Give me a second.
            $endgroup$
            – Martin Argerami
            Dec 19 '18 at 4:56










          • $begingroup$
            You can do it as I did, or you find that $C=0$. In any case, the limit is zero.
            $endgroup$
            – Martin Argerami
            Dec 19 '18 at 4:59
















          0












          $begingroup$

          You have
          $$
          v(t)=lntfrac23+int_0^tfrac1{u^2+5u+6},du=lntfrac23+lnfrac{t+2}{t+3}+lntfrac32,
          $$
          where the $ln tfrac32$ comes from evaluating the antiderivative at $0$. Then
          $$
          lim_{ttoinfty}v(t)=ln tfrac23+ln tfrac32=0.
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I didn't understand a thing you did. Doesn't it tell us that the velocity is initially ln(2/3) so that we can use that as an initial condition to plug t = 0 and solve for C, which I did already and got C = 0. So why did you do that additional step.
            $endgroup$
            – krauser126
            Dec 19 '18 at 4:52










          • $begingroup$
            Now I see that I misread something. Give me a second.
            $endgroup$
            – Martin Argerami
            Dec 19 '18 at 4:56










          • $begingroup$
            You can do it as I did, or you find that $C=0$. In any case, the limit is zero.
            $endgroup$
            – Martin Argerami
            Dec 19 '18 at 4:59














          0












          0








          0





          $begingroup$

          You have
          $$
          v(t)=lntfrac23+int_0^tfrac1{u^2+5u+6},du=lntfrac23+lnfrac{t+2}{t+3}+lntfrac32,
          $$
          where the $ln tfrac32$ comes from evaluating the antiderivative at $0$. Then
          $$
          lim_{ttoinfty}v(t)=ln tfrac23+ln tfrac32=0.
          $$






          share|cite|improve this answer











          $endgroup$



          You have
          $$
          v(t)=lntfrac23+int_0^tfrac1{u^2+5u+6},du=lntfrac23+lnfrac{t+2}{t+3}+lntfrac32,
          $$
          where the $ln tfrac32$ comes from evaluating the antiderivative at $0$. Then
          $$
          lim_{ttoinfty}v(t)=ln tfrac23+ln tfrac32=0.
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 19 '18 at 4:58

























          answered Dec 19 '18 at 4:49









          Martin ArgeramiMartin Argerami

          127k1182183




          127k1182183












          • $begingroup$
            I didn't understand a thing you did. Doesn't it tell us that the velocity is initially ln(2/3) so that we can use that as an initial condition to plug t = 0 and solve for C, which I did already and got C = 0. So why did you do that additional step.
            $endgroup$
            – krauser126
            Dec 19 '18 at 4:52










          • $begingroup$
            Now I see that I misread something. Give me a second.
            $endgroup$
            – Martin Argerami
            Dec 19 '18 at 4:56










          • $begingroup$
            You can do it as I did, or you find that $C=0$. In any case, the limit is zero.
            $endgroup$
            – Martin Argerami
            Dec 19 '18 at 4:59


















          • $begingroup$
            I didn't understand a thing you did. Doesn't it tell us that the velocity is initially ln(2/3) so that we can use that as an initial condition to plug t = 0 and solve for C, which I did already and got C = 0. So why did you do that additional step.
            $endgroup$
            – krauser126
            Dec 19 '18 at 4:52










          • $begingroup$
            Now I see that I misread something. Give me a second.
            $endgroup$
            – Martin Argerami
            Dec 19 '18 at 4:56










          • $begingroup$
            You can do it as I did, or you find that $C=0$. In any case, the limit is zero.
            $endgroup$
            – Martin Argerami
            Dec 19 '18 at 4:59
















          $begingroup$
          I didn't understand a thing you did. Doesn't it tell us that the velocity is initially ln(2/3) so that we can use that as an initial condition to plug t = 0 and solve for C, which I did already and got C = 0. So why did you do that additional step.
          $endgroup$
          – krauser126
          Dec 19 '18 at 4:52




          $begingroup$
          I didn't understand a thing you did. Doesn't it tell us that the velocity is initially ln(2/3) so that we can use that as an initial condition to plug t = 0 and solve for C, which I did already and got C = 0. So why did you do that additional step.
          $endgroup$
          – krauser126
          Dec 19 '18 at 4:52












          $begingroup$
          Now I see that I misread something. Give me a second.
          $endgroup$
          – Martin Argerami
          Dec 19 '18 at 4:56




          $begingroup$
          Now I see that I misread something. Give me a second.
          $endgroup$
          – Martin Argerami
          Dec 19 '18 at 4:56












          $begingroup$
          You can do it as I did, or you find that $C=0$. In any case, the limit is zero.
          $endgroup$
          – Martin Argerami
          Dec 19 '18 at 4:59




          $begingroup$
          You can do it as I did, or you find that $C=0$. In any case, the limit is zero.
          $endgroup$
          – Martin Argerami
          Dec 19 '18 at 4:59


















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