Solve this high...

Discussion in 'General' started by Borstal, Sep 28, 2009.

  1. Alright so my buddy is over at carleton uni and he sent me this question he has been stuck on for sometime and I have been trying to help him but im blazed as fuck so I'd thought I would post it here so we can all be confused.

    An object, moving at constant acceleration covers two equal, consecutive, 11 m long segments, the first one in t1 = 0.93 s and the second one in t2 = 2.37s . What was the speed of the object (in m/s) at the beginning of the first segment?
     
  2. fuck if i know
     
  3. #4 Jakigi, Sep 28, 2009
    Last edited by a moderator: Sep 28, 2009
    D=V(0)t+1/2(a)((t)^2)
    11m=0+1/2(a)((.93s)^2)

    a=25.4m/s^2

    V=V(0)+(a)(t)
    V=0+(25.4)(.93s)

    V (at the end of the first segment)=23.62m/s

    (edit 1: wait... beginning of the first segment? 0m/s.... because no time had passed. V=v(0)+at... regardless of the acceleration, if no time had passed -->0m/s)

    (edit 2: wait... i assumed it started at rest. did it? if not... meh)
     
  4. respond fucker! this is killing me...
     
  5. first of all, its undergoing constant deceleration.

    yeah there needs to be more information to accurately solve this problem..
     
  6. Oh yeah...

    Hmm...

    Fuck physics...
     
  7. i got 10.2 m/s.. what did anybody else get
     
  8. i looked at this and was like damn im way 2 stoned 2 even turn that into numbers:smoking:
     
  9. #10 Jakigi, Sep 28, 2009
    Last edited by a moderator: Sep 29, 2009
    V1(1.44s)+1/2(a(t^2))=V0(.93s)+1/2(a(t^2))

    Substitute V1 --> V0+at

    (V0+(.93s)(a))*1.44s)+1/2(a)(1.44s^2)=V0(.93s)+1/2(a)(.93s^2)

    (1.44V0+1.34a)+(1.04a)=(.93V0)+(0.43a)

    Simplified --> -3.82a=V0

    11=-3.82a(.93s)+1/2(a)(.93s^2)

    11=-3.12a

    a= -3.53 m/s^2

    Plug back into --> -3.82a=V0

    V0=13.47m/s

    Vf=13.47+(-3.53)(.93s)

    Vf=10.18 m/s

    FUCK YOU OP!!!!!!!!!!!!!!:smoking:
     
  10. i dont even remember what i did but i pretty much got it if you rounded your answer of 10.18m/s... uh it was real simple hit us with some harder questions. btw im not even in college yet and can solve that. booo
     
  11. damn im in algebra, and i thought that shit was hard.
     
  12. lol quit gettin us to do your math homework op lol jk
     
  13. no such thing as deceleration in physics.
    its accelerating negatively/ in the other direction.

    and that problem is solveable.
    I'm too tired to do it but it's not all that hard.

    edit: is the speed for the second segment even necessary information?
     


  14. oh god....fuck if i know too...im still on regular multiplications and im a super senior :(
     



  15. yeah fuck physics
     
  16. physics is awesome
     
  17. I'm a bit lost.

    If it's moving at constant acceleration then why does the 2nd segment take longer than the first segment?

    It's been wayy too many years since I did math like that, but I'm pretty sure the 2nd segment should be quicker than the 1st segment at constant acceleration.
     
  18. constant *negative* acceleration.

    that means it's slowing down and didn't start at rest...

    im pretty sure i did it correctly.... blknazn and I got 10.2 m/s.

    but, still waiting for the op to respond.
     
  19. ok....im gonna figure this out hella quick....hold on...
     

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