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Simple Problem, But I'm a 9th Grader :P

  1. Sep 26, 2004 #1
    ok im in 9th grade and im taking honors physics. were not very far into it, juss like simple motion u kno? ive tried to take in a lot at once recently bc ive been sick, so i went and read ch. 3-5 in my book...well theres sections i cant find again and stuff and i need help with the following problem:
    Show mathematically that an object can have a negative position but a positive velocity.
    I had this idea that it can go to like the left at 45 m/s or something...but im not sure :P help!
     
  2. jcsd
  3. Sep 26, 2004 #2
    omg theres like 3 more too...IM SOOOO CONFUSED! :cry:
     
  4. Sep 26, 2004 #3

    Gokul43201

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    First draw a picture. Show the position axis with the origin. Mark a spot where the object can be for it to have a negative position. Now use an arrow to represent the velocity vector. How must this arrow point for the velocity to be positive ?

    In the future, post this in the Homework Help section.

    (would someone move this ?)
     
  5. Sep 26, 2004 #4
    oh like to the right it just has to start froma negative position rite?
     
  6. Sep 26, 2004 #5

    tony873004

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    position = original position + velocity * time

    There's no reason that position or original position can't be negative. They're not under a squre root symbol. Given enough time, position will cross 0 and become positive. And position is just a relative thing anyway. A car travelling east and is 1 mile west of mile marker 2 could be said to have a position of -1 mile relative to mile marker 2, or +1 mile relative to mile marker 0.

    But I don't know how to express that mathematically.
     
  7. Sep 26, 2004 #6
    im not sure if it means an equation or a graph or what...omg im gona get an F :(
     
  8. Sep 26, 2004 #7

    Gokul43201

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    That's right.

    Consider x(t=0s) = -10m (this is a negative position)
    and x(t=1s) = -5m (also negative)

    The v = (x1-x0)/t = ?
     
  9. Sep 27, 2004 #8
    Okay wait.. v=(x1-xo)/t =?
    v=(-10+5)/1s ????
    v=-5m/s? is that rite?
     
  10. Sep 27, 2004 #9
    YES! I GOT IT!...okay it juss kinda clicked lol...but i have a small question...okay heres the problem:
    A stunt car is driven along a flat train car. The stunt car is moving toward the engine of the trani. how would you calculate the velocity of the stunt car relative to Earth?
    I put. . .'You would find the velocity of the ____ then the velocity of the stunt car. Then subtract using V2-V1...' Do i find the velocity of the train? And if it is the train, then is that V2?
     
  11. Sep 27, 2004 #10

    tony873004

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    Some trains have more than one engine. Assuming that it meant that the car is moving towards the front of the train, you would add the train's velocity to the car's velocity.
     
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