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Avg speed

  1. Sep 7, 2009 #1
    1. The problem statement, all variables and given/known data
    A car travels up a hill at a constant speed of 32 km/h and returns down the hill at a constant speed of 66 km/h. Calculate the average speed (in km/h) for the round trip.


    2. Relevant equations

    average speed= total distance/ delta time is the only equation i know for it

    3. The attempt at a solution

    there is no time, or actual distance given in this question, so i got confused and just tried to find the average from adding up the two speeds.. obviously that didn't work- this seems like a really simple problem, so where am i going wrong?
     
  2. jcsd
  3. Sep 7, 2009 #2
    I don't think your doing anything wrong. I think the problem is missing the time information.

    Thanks
    Matt
     
  4. Sep 7, 2009 #3
    but... it's online homework. that was just the given problem :/
     
  5. Sep 7, 2009 #4
    Well, I have solved many dynamics problems and this problem is missing the information needed to solve it. Can you contact the teacher/instructor who posted it?

    Thanks
    Matt
     
  6. Sep 7, 2009 #5

    Vanadium 50

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    CFDFEAGURU is, I fear, leading you astray. There is no missing fact, although there is a very important word, "returns". That means the distance up the hill is the same as the distance down the hill.
     
  7. Sep 7, 2009 #6
    Whoops LOL. I missed that word.

    Sorry to the OP.

    Thanks
    Matt
     
  8. Sep 7, 2009 #7
    But.. it's average speed, not velocity.. so, direction doesn't matter, i thought? and you can't just average the speeds they give you.. and there's no time given :/
     
  9. Sep 7, 2009 #8

    Vanadium 50

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    Suppose the distance is 1km. Work out average speed. 2km? 5km? See a pattern?

    Now try x km.
     
  10. Sep 7, 2009 #9
    so.. like, suppose 1 km is the distance

    so do 32 km/hr / 1 km = 32 hr^-1 ,etc?

    then what? :/ i'm sorry, i just really need this to be actually taught to me..
     
  11. Sep 7, 2009 #10
    Well, we know that
    [tex]s=\frac{d}{t}[/tex]
    [tex]s=\frac{d_1+d_2}{t_1+t_2}[/tex]
    d1 and t1 are for the first have of the trip @ 32 km/h and d2 and t2 are for the other half at 66km/h.
    Now you need to find what each of those are equal to.
     
  12. Sep 7, 2009 #11
    but, how..
     
  13. Sep 7, 2009 #12
    [tex]d_1=d_2=x[/tex]
    [tex]v_1 = \frac{x}{t_1}[/tex]
    [tex]v_2 = x/t_2[/tex]
    Solve for the t's and plug into the speed equation in my last post. Then do some algebra.
     
  14. Sep 7, 2009 #13
    so i have s= 2x/(x/32 + x/66)

    but.. if we don't know what s is and we don't know what x is
    what comes next?
     
  15. Sep 7, 2009 #14
    [tex]s=\frac{2x}{\frac{x}{32}+\frac{x}{66}}[/tex]
    If you find a common denominator for the lowed half of the equation the x's will cancel out.
     
  16. Sep 7, 2009 #15
    yes, of course you're right

    thank you very much for your help, i really appreciate it
     
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