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Homework Help: Desperately need help -_-

  1. Oct 3, 2007 #1
    *SOLVED*Desperately need help -_-


    1. The problem statement, all variables and given/known data

    You and a friend each drive 50 km. You travel at 90km/h; your friend travels at 95km/h. How long will your friend wait for you at the end of the trip?

    Df(Displacement Final): 50km
    Di(Displacement Initial): 0
    Vme(My Avg Velocity): 90km/h
    Vfriend(etc.): 95km/h

    2. Relevant equations

    I don't know! It's like I know I should know them but I can't seem to find anything in the chapter -_-

    3. The attempt at a solution

    I attempted first to break this down(since all the equations I know use seconds and meters) via dimensional analysis and this is how it turned out(note: I rounded one of the numbers)

    Code (Text):
    90km     1h     90km     1m    90km        ended up w/ 0.025km/s
    ----- x ----- = ----  x ---- = -----        for the friend I got 0.026km/s
      1h    60m     60m     60s    3600s
    then I did the stupid thing making it m instead of km, resulting in this:

    90km=0.000025m 95km=0.000026m

    But I don't really know what to do from here; I mean I know that I need to make those conversions but from here I don't know what formula to use!

    Were my conversions wrong? Unnecessary?

    What formula do I need! Someone help, I have a quiz on this stuff tomorrow -_-

    Thanks in advance!
    Last edited: Oct 3, 2007
  2. jcsd
  3. Oct 3, 2007 #2
    Hold on I think I may have just figured it out -_-

    Stupid me remember that T=D/V

    Code (Text):

     50km     .05m       .05m
    ------ = ------- = ---------- = 2000s
    90km/h   .09m/h   .000025m/s

     50km     .05m       .05m
    ------ = ------- = ---------- = 1923.0769230769s
    95km/h   .095m/h  .000026m/s
    THEREFORE my friend would have to wait roughly 77s

  4. Oct 3, 2007 #3
    You don't need to do any conversion. Well, to clean things up, you might want to, but it doesn't seem like the problem doesn't require to.

    Basically, the main idea here is to find the time it takes for you and your friend to get to the destination. You know velocity = distance / time, so time = distance / velocity. Plug in the individual velocities, and you should find the time in hours. Subtract the time required for you to get to the destination by the time for the friend to get to the destination, and that's how long your friend must wait for you.
    Last edited: Oct 3, 2007
  5. Oct 3, 2007 #4

    Oh you figured out even before I submitted my response lol.

    Check your answer for the second calculation. I got a number smaller than that.
  6. Oct 3, 2007 #5
    wait what? The one for .095m?


    I see what happened; I just took the rounded number my calc. spit out;

    the full number is like 2.6388888888889 x 10-5

    Resulting in the answer of 1894.7368421053
    Last edited: Oct 3, 2007
  7. Oct 3, 2007 #6
    Crap, you are right, Velocity = Distance / Time, So Time = Distance / Velocity.

    Again, check your answer on 2nd part.
  8. Oct 3, 2007 #7
    Yes. You made a numerical mistake.
  9. Oct 3, 2007 #8
    So you would figure out how long it takes your friend to do the trip

    t_friend = 50km/95km/h

    and how long it takes you to do the trip

    t_you = 50km/90km/h

    If you subtract your time from his time then that tell you how long he was waiting in hours, which happens to be .03 of an hour, do the math to make sure I'm not pulling your leg.

    I leave it to you to figure out how many seconds that is.
    Last edited: Oct 3, 2007
  10. Oct 3, 2007 #9
    here's the edit I wrote -_-

  11. Oct 3, 2007 #10
    ^^^ Still not right.
  12. Oct 3, 2007 #11
    Um -_-

    So when I subtract 1894.7368421053 from 2000 I come out w/ 105.2631578947s

    which is like 1 minute and 45 seconds or something

    how do you come out w/ 6 minutes??
  13. Oct 3, 2007 #12
    You should also really check that your answer makes sense. 1894 seconds would roughly be half an hour, that doesn't make sense that he would wait that long when you were going almost the same speed.

    Oh, I see what you have as 1894. Nvm, let me look at what else you did.
  14. Oct 3, 2007 #13
    nonononono 1894 s is how long his trip took -_-

    the 2000s and 1894s were how long each of our trips took -_-

  15. Oct 3, 2007 #14
    Yeah, this is right. I'm not sure why this would be wrong, this is what I got.
  16. Oct 3, 2007 #15
    I think I have solved this one; It wasn't that hard once I found the right formula -_-

    Thanks to both of you guys, you really saved my butt ^_^

    I have a bunch of other problems to do though; She was out and gave us a bunch of work and I wasn't there when it was given so I have to do like 25 more...I've only gotten through the first 5...counting this one -_-
  17. Oct 3, 2007 #16
    Because google calculator lied to me. You got it right.
  18. Oct 3, 2007 #17
    Lol thanks dude! I didn't mean like "haha omfg how did you get 6 minutes!!!" I was just thinking I did something wrong -_-

    Also, apologies for my crappy punctuation; I am usually a lot better about it, but today I really don't give a crap...at least not right now ;-)
  19. Oct 3, 2007 #18
    i got 105.26 seconds. approx 1 min 45 seconds.

    my method: (50km/90km)*60 minutes = 33.333 mins (this is the time the first guy takes to complete the 50km.

    (50km/95km)*60 minutes = 31.58 mins.

    33.33-31.58 = approx. 1 min 45.
  20. Oct 3, 2007 #19
    Awesome, as did I;

    One more question, for the next problem -_-

    How do you figure Instantaneous Velocity from a point on a graph?

    A. Car B at 2.0s
    d(m) = 1.5

    How do I snag this? You are supposed to take the slope of the line at that point right?
  21. Oct 3, 2007 #20
    The easiest way is the way I first suggested, given you don't trust google to do it right.

    Find the discrepancy in hours and then multiply by 3600s/h. Discrepancy was about .03hrs, you should keep it in irrational form to get the right answer ((50/90)-(50/95))hrs * 3600s/h.

    Yes the answer is about 100s.
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