my friend has a physics problem..

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  • robfantasy
    Confirmed User
    • Jun 2002
    • 6445

    #1

    my friend has a physics problem..

    driving along a freeway, you notice that it takes a time t to go from one mile marker to the next. When you increase your speed by 8 mi/h , the time to go one mile decreases by 10 s . What was your original speed?

    how do u get this answer..
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  • SicChild
    Confirmed User
    • Mar 2003
    • 365

    #2
    I think it's a system problem.

    v = d / t

    so,

    t = 1 (1 mile) / v
    and
    t - 10 = 1 / (v + 8)

    isolate a variable and solve for it...

    Comment

    • ADL Colin
      Too lazy to set a custom title
      • Feb 2001
      • 11929

      #3
      v1 = first velocity
      v2= second velocity
      d1- first distance
      d2 = second distance

      d1 = d2 = 1
      Since d = v/t

      v1*t = v2*t2 = 1

      Using:
      v2 = v1 +8
      and t2 = t = 10

      (v1 + 8)(t-10) = 1
      10v1 ? 8t + 80 = 0
      10v1^2 + 80v1 ? 8 = 0

      Using quadratic formula with a=10, b=80, c = -8

      You get v1 = ~ .09878 mph which you can verify is correct by :::

      In the first mile .09878 mph means you finish in 10.12 seconds. Then in the second mile you travel at 8.09875 mph which means you finish in .12 seconds


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      • peteinoz
        Confirmed User
        • Nov 2002
        • 183

        #4
        Originally posted by Colin
        v1 = first velocity
        v2= second velocity
        d1- first distance
        d2 = second distance

        d1 = d2 = 1
        Since d = v/t

        v1*t = v2*t2 = 1

        Using:
        v2 = v1 +8
        and t2 = t = 10

        (v1 + 8)(t-10) = 1
        10v1 ? 8t + 80 = 0
        10v1^2 + 80v1 ? 8 = 0

        Using quadratic formula with a=10, b=80, c = -8

        You get v1 = ~ .09878 mph which you can verify is correct by :::

        In the first mile .09878 mph means you finish in 10.12 seconds. Then in the second mile you travel at 8.09875 mph which means you finish in .12 seconds
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        • Holly
          Too lazy to set a custom title
          • Jun 2003
          • 10017

          #5
          Originally posted by Colin
          v1 = first velocity
          v2= second velocity
          d1- first distance
          d2 = second distance

          d1 = d2 = 1
          Since d = v/t

          v1*t = v2*t2 = 1

          Using:
          v2 = v1 +8
          and t2 = t = 10

          (v1 + 8)(t-10) = 1
          10v1 ? 8t + 80 = 0
          10v1^2 + 80v1 ? 8 = 0

          Using quadratic formula with a=10, b=80, c = -8

          You get v1 = ~ .09878 mph which you can verify is correct by :::

          In the first mile .09878 mph means you finish in 10.12 seconds. Then in the second mile you travel at 8.09875 mph which means you finish in .12 seconds
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          Comment

          • freeadultcontent
            Confirmed User
            • Oct 2002
            • 9976

            #6
            Thank you for reminding me I suck at all math asside from accounting.

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            • KDizzla
              Confirmed User
              • Sep 2002
              • 1795

              #7
              Originally posted by freeadultcontent
              Thank you for reminding me I suck at all math asside from accounting.
              Math has always been a bad to me.
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              Comment

              • Holly
                Too lazy to set a custom title
                • Jun 2003
                • 10017

                #8
                Originally posted by KDizzla
                Math has always been a bad to me.
                I just noticed that in the Jessica Lynch thread.
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                • ADL_Christy
                  Confirmed User
                  • Mar 2004
                  • 304

                  #9
                  I knew Colin would have the answer to this!!!!

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                  • ADL Colin
                    Too lazy to set a custom title
                    • Feb 2001
                    • 11929

                    #10
                    Originally posted by ADL_Christy
                    I knew Colin would have the answer to this!!!!
                    Hey Christy! :-)


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                    • hjnet
                      Confirmed User
                      • May 2002
                      • 3815

                      #11
                      Originally posted by Colin
                      v1 = first velocity
                      v2= second velocity
                      d1- first distance
                      d2 = second distance

                      d1 = d2 = 1
                      Since d = v/t

                      v1*t = v2*t2 = 1

                      Using:
                      v2 = v1 +8
                      and t2 = t = 10

                      (v1 + 8)(t-10) = 1
                      10v1 ? 8t + 80 = 0
                      10v1^2 + 80v1 ? 8 = 0

                      Using quadratic formula with a=10, b=80, c = -8

                      You get v1 = ~ .09878 mph which you can verify is correct by :::

                      In the first mile .09878 mph means you finish in 10.12 seconds. Then in the second mile you travel at 8.09875 mph which means you finish in .12 seconds

                      I don't wanna be the ass here, but I think you mixed up seconds and hours somewhere.

                      At the beginning he drove with 50mph (and needed 72 sec for 1mile), then he increased his speed to 58mph and therefore only needed 62 sec for a mile

                      Comment

                      • ADL Colin
                        Too lazy to set a custom title
                        • Feb 2001
                        • 11929

                        #12
                        Originally posted by hjnet
                        I don't wanna be the ass here, but I think you mixed up seconds and hours somewhere.

                        At the beginning he drove with 50mph (and needed 72 sec for 1mile), then he increased his speed to 58mph and therefore only needed 62 sec for a mile

                        Haha. Oh, shit. You're right. Forgot to convert the units.


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