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Too Many Unknowns

  1. Oct 10, 2004 #1
    Hello, everyone. I'm new here, drawn by a velocity problem that's had me going in circles for two days. If someone could tell me how to set this up/proceed with solving, that'd be great.

    "On a car trip, you want to have an average velocity of 90km/h. Over the first half of the trip, you averaged only 48 km/h. What must your velocity be in the second half to obtain an overall average velocity of 90 km/h? Note that the velocities are based on half the distance, not half the time."

    So far, I've slipped up so many times that I eventually came out with
    "48 km/h = 48 km/h"

    Our physics teacher is a nice guy, but he hasn't taught us well enough on dealing with many unknowns for me to be able to solve this without help.

  2. jcsd
  3. Oct 10, 2004 #2


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    Let's start slow: what are the unknowns?
  4. Oct 10, 2004 #3
    hmm you should probably show some work


    I am not sure what the problem is since it is half the distance

    average equals the total of the terms divided by the number of terms

    you have

    [tex] \frac {48+x}{{2}}=90 [/tex]
    [tex] 48+x=180 [/tex]
    [tex] x=180-48=132 [/tex]

    Note to admin: if you feel that this solution should not be posted feel free to remove but it seemed like it may help to see general form for additonal problems.
  5. Oct 10, 2004 #4


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    Oh, heck, I'll leave it in, if only because it won't help the poor guy.
    It's completely wrong! You just can't average speeds over different times like that.

    To Afro_Akuma: If you are concerned that the distance isn't given, maybe it doesn't matter! If you assume the total distance is 200 km, what answer do you get? If you assume the total distance is 2000 km, what answer do you get?

    (By the way, I am assuming that "the first half of the trip" means half the distance, not half the time.)
  6. Oct 10, 2004 #5
    wow I suck at reading
    sorry i did half the time...
  7. Oct 10, 2004 #6
    Answer= 720 km/hr

    thats what I got
    can anyone confirm?
  8. Oct 10, 2004 #7

    here is how i did it

    for a basic average of speed lets say you had the following

    3 m/s for 5 sec
    4 m/s for 8 sec

    to find average you would take (15+32)/(13)

    you realize you take the distances and divide it by total time

    so lets test some different values for x

    First I tested 2000 total

    since we know each section is 1000 and we need the

    [tex] \frac{rate_1*time_1+rate_2*time_2}{time_1+time_2}=90 [/tex]

    however since rate*time equals distance than we can make the equation

    [tex] \frac{distance_1+distance_2}{time_1+time_2}=90 [/tex]

    so lets subsititue for 2000

    sense distance1=distance2=1000
    time_1= 1000/48
    [tex] \frac{2000}{{\frac{1000}{48}+time_2}}=90 [/tex]


    [tex] 2000=90(\frac{1000}{48}+time_2) [/tex]

    therefore time 2 equals



    since [tex] 1000=(125/90)*rate [/tex]
    rate equals 720

    Then try it with other values to confirm you will get 720 again
    here is short
    using total distance equal to 96

    [tex] \frac{96}{1+x}=90 [/tex]
    6/90 = x
    48= (6/90)*rate
    Last edited: Oct 10, 2004
  9. Oct 10, 2004 #8
    Sorry for answering prematurely
    I honestly just misread the question.... thanks for pointing it out though it was fun to do. I ended up scribbling on the back of my paper that had my essay questions for MIT lol....
  10. Oct 11, 2004 #9
    Unknowns: Time (any), distance/displacement (any)
    Known: No acceleration, 48 km/h velocity over half the total distance, average velocity of 90 km/h.

    Yes, the "first half of the trip" meant by distance. My problem is that I can't just set an appropriate time/ditance and substitute it into the problem. The teacher's a big algebra fan. Variables all the way through to the end, no numbers that weren't sourced directly from the problem.

    As for showing my work, much of it is completely useless. I've done things such as:

    ------- = 90 km/h


    ------ = 48 km/h


    ------ = 96 km/h

    Was I on ther right track?
  11. Oct 11, 2004 #10


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    You are starting off okay. What you really want to calculate are the times.

    Let d be the total distance, as you did. Then the person drives half the distance at 48 km/h so (1/2)d/t1= 48, d/t1= 96 (as you have). The d= 96t1 so t1= d/96. Let "v" be the speed for the second half and t2 the time. Now (1/2)d/t2= v so d= 2v t2 and t2= d/2t2. The average speed for the entire trip must be 90 km/h so, letting t be the entire time, d/t= 90 or t= d/90.

    t= t1+ t2. Put the values we have, t= d/90, t1= d/96, and t2= d/2t2 into that and solve for v. You can watch d magically disappear from the equation!
  12. Oct 11, 2004 #11
    was I right in my second explantion with an answer of 720 km/hr
  13. Oct 11, 2004 #12
    This has been a great help to me, but I'm confused on one point:
    In t2 = d/2t2, where did 2t2 come from?
  14. Oct 11, 2004 #13
    I don't understand this lst part. Should it be t2 = d/2v?
  15. Oct 11, 2004 #14
    Here is an algerbraic explantion

    [tex] \frac{d}{time_1+time_2}=90 [/tex]
    [tex] time=distance/rate [/tex]

    Total Distance (d)
    ------------------- =90
    [tex] \frac{.5d}{48} + \frac{.5d}{rate_2} [/tex]

    get common den
    [tex] \frac {d}{.5d*rate_2+24*d}=90 [/tex]
    [tex]48*rate_2 [/tex]
    flip simplify

    [tex] \frac {d*48*rate_2}{d(.5rate_2+24}=90 [/tex]

    [tex] 48*rate_2=45*rate_2+2160 [/tex]

    [tex] 3*rate_2=2160 [/tex]

    [tex] rate_2=720 km/h [/tex]

    Last edited: Oct 11, 2004
  16. Oct 11, 2004 #15
    Where did rate come in?
  17. Oct 11, 2004 #16
    Actually, I figured out that rate = velocity. But could you explain the "Flip Simplify" step?
    It just threw me completely how you arrived at that.
  18. Oct 11, 2004 #17


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    A complex fraction:

    can be flipped as the denominator of the denominator is the numerator.

  19. Oct 11, 2004 #18
    Thanks! That helps a lot!
  20. Oct 11, 2004 #19
    sorry I couldn't figure out how to do triple fractions with latex, but Ba was right in flipping fractions--- for simplifying I just multiply stuff out and cancel
  21. Oct 11, 2004 #20
    you have to be carful somtimes though because




    what Ba simplified which is what I had was

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