Calculus Subway Problem

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A subway train travels over a distance $s$ in $t$ seconds. It starts from rest (zero velocity) and ends at rest. In the first part of its journey, it moves with constant acceleration $f$ and in the second part with constant deceleration (negative deceleration) $r$. Show that $$s = fracfracfrt^2f + r2$$




I tried doing this multiple times but to no avail. I used $x$ to represent the number of seconds traveled with acceleration $f$ and then $t - x$ to represent the number of seconds decelerating. I'm still not sure how to do it. Any help would be appreciated.



Thanks.




calculus word-problem




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    A subway train travels over a distance $s$ in $t$ seconds. It starts from rest (zero velocity) and ends at rest. In the first part of its journey, it moves with constant acceleration $f$ and in the second part with constant deceleration (negative deceleration) $r$. Show that $$s = fracfracfrt^2f + r2$$




    I tried doing this multiple times but to no avail. I used $x$ to represent the number of seconds traveled with acceleration $f$ and then $t - x$ to represent the number of seconds decelerating. I'm still not sure how to do it. Any help would be appreciated.



    Thanks.




    calculus word-problem




    share|cite|improve this question





















      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite












      A subway train travels over a distance $s$ in $t$ seconds. It starts from rest (zero velocity) and ends at rest. In the first part of its journey, it moves with constant acceleration $f$ and in the second part with constant deceleration (negative deceleration) $r$. Show that $$s = fracfracfrt^2f + r2$$




      I tried doing this multiple times but to no avail. I used $x$ to represent the number of seconds traveled with acceleration $f$ and then $t - x$ to represent the number of seconds decelerating. I'm still not sure how to do it. Any help would be appreciated.



      Thanks.




      calculus word-problem




      share|cite|improve this question












      A subway train travels over a distance $s$ in $t$ seconds. It starts from rest (zero velocity) and ends at rest. In the first part of its journey, it moves with constant acceleration $f$ and in the second part with constant deceleration (negative deceleration) $r$. Show that $$s = fracfracfrt^2f + r2$$




      I tried doing this multiple times but to no avail. I used $x$ to represent the number of seconds traveled with acceleration $f$ and then $t - x$ to represent the number of seconds decelerating. I'm still not sure how to do it. Any help would be appreciated.



      Thanks.





      calculus word-problem





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      asked Jul 28 at 22:32









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          The area under the curve equals the distance traveled because integratingis really just adding up every value of the curve. The slope of the left side it $f$ and the slope of the right side is $-r$.
          $$s=mboxarea=frac12v_mt$$
          Now lets find $v_m$ in terms of $f$ and $r$



          Note that the slope is rise over run. $f=dfracv_mx$ and $r=dfracv_mt-x$



          So we need to eliminate $x$ and find $v_m$$$x=fracv_mf,x=t-fracv_mr$$$$fracv_mf=t-fracvMr$$$$rv_m=frt-fv_m$$$$v_m=fracfrf+rt$$$$s=frac12left[fracfrf+rtright]t$$$$s=left[fracfrf+rright]fract^22$$$$s=fracfracfrt^2f+r2$$






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            enter image description here



            The area under the curve equals the distance traveled because integratingis really just adding up every value of the curve. The slope of the left side it $f$ and the slope of the right side is $-r$.
            $$s=mboxarea=frac12v_mt$$
            Now lets find $v_m$ in terms of $f$ and $r$



            Note that the slope is rise over run. $f=dfracv_mx$ and $r=dfracv_mt-x$



            So we need to eliminate $x$ and find $v_m$$$x=fracv_mf,x=t-fracv_mr$$$$fracv_mf=t-fracvMr$$$$rv_m=frt-fv_m$$$$v_m=fracfrf+rt$$$$s=frac12left[fracfrf+rtright]t$$$$s=left[fracfrf+rright]fract^22$$$$s=fracfracfrt^2f+r2$$






            share|cite|improve this answer

























              up vote
              1
              down vote



              accepted










              enter image description here



              The area under the curve equals the distance traveled because integratingis really just adding up every value of the curve. The slope of the left side it $f$ and the slope of the right side is $-r$.
              $$s=mboxarea=frac12v_mt$$
              Now lets find $v_m$ in terms of $f$ and $r$



              Note that the slope is rise over run. $f=dfracv_mx$ and $r=dfracv_mt-x$



              So we need to eliminate $x$ and find $v_m$$$x=fracv_mf,x=t-fracv_mr$$$$fracv_mf=t-fracvMr$$$$rv_m=frt-fv_m$$$$v_m=fracfrf+rt$$$$s=frac12left[fracfrf+rtright]t$$$$s=left[fracfrf+rright]fract^22$$$$s=fracfracfrt^2f+r2$$






              share|cite|improve this answer























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                enter image description here



                The area under the curve equals the distance traveled because integratingis really just adding up every value of the curve. The slope of the left side it $f$ and the slope of the right side is $-r$.
                $$s=mboxarea=frac12v_mt$$
                Now lets find $v_m$ in terms of $f$ and $r$



                Note that the slope is rise over run. $f=dfracv_mx$ and $r=dfracv_mt-x$



                So we need to eliminate $x$ and find $v_m$$$x=fracv_mf,x=t-fracv_mr$$$$fracv_mf=t-fracvMr$$$$rv_m=frt-fv_m$$$$v_m=fracfrf+rt$$$$s=frac12left[fracfrf+rtright]t$$$$s=left[fracfrf+rright]fract^22$$$$s=fracfracfrt^2f+r2$$






                share|cite|improve this answer













                enter image description here



                The area under the curve equals the distance traveled because integratingis really just adding up every value of the curve. The slope of the left side it $f$ and the slope of the right side is $-r$.
                $$s=mboxarea=frac12v_mt$$
                Now lets find $v_m$ in terms of $f$ and $r$



                Note that the slope is rise over run. $f=dfracv_mx$ and $r=dfracv_mt-x$



                So we need to eliminate $x$ and find $v_m$$$x=fracv_mf,x=t-fracv_mr$$$$fracv_mf=t-fracvMr$$$$rv_m=frt-fv_m$$$$v_m=fracfrf+rt$$$$s=frac12left[fracfrf+rtright]t$$$$s=left[fracfrf+rright]fract^22$$$$s=fracfracfrt^2f+r2$$







                share|cite|improve this answer













                share|cite|improve this answer



                share|cite|improve this answer











                answered Jul 28 at 22:44









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