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Finding the distance in an acceleration problem?

  1. Mar 10, 2010 #1
    1. The problem statement, all variables and given/known data
    What distance is required to stop a car going 90km/h if the car can stop with an acceleration of -6.0m/s2?


    2. Relevant equations
    a = v2-v1/t
    d = (v1+v2/2)t


    3. The attempt at a solution
    I'm thinking first we have to solve for t. Which would be the speed divided by acceleration, which is -0.36 km/h^2 over 80km/h which is 0.0045 hours.

    Now to solve for distance, it would be 80km/2 x 0.0045

    Which leaves me with 0.18 which is completely wrong because the answer is supposed to be 41 m.

    Help please? Thank you.
     
  2. jcsd
  3. Mar 10, 2010 #2
    Oops speed over acceleration is actually 80km/h over -0.36km/h, but that still gives me the wrong answer..
     
  4. Mar 10, 2010 #3
    Check your equation for D again.
     
  5. Mar 10, 2010 #4
    It's right ?

    D = (V1 + v2/2) x T
     
  6. Mar 10, 2010 #5
    1. Is it 90 or 80 km/h?
    2. your formula should be [tex]\Delta d=\frac{v_{1}+v_{2}}{2}\Delta t[/tex]
     
  7. Mar 10, 2010 #6
    It says 80km on my sheet.
    But yeah, that is the formula I am using.
    (I do v1+v2 first, and then divide it by 2)
     
  8. Mar 10, 2010 #7
    I think I see your mistake. How did you convert [tex]m/s^{2}[/tex] to [tex]km/h^{2}[/tex]
     
  9. Mar 10, 2010 #8
    Oh sorry for writing 90 in the first post, that was a typo. Should be 80km.

    Okay so it's 80km/h.
    I converted that -5.0m/s into km/^s by doing this:

    -5/1000
    = - 0.005

    Then I multiplied this by 60..

    But now I'm getting -0.3. x_x
    Ahh, can anyone thoroughly help me? This is so frustrating! I am so lost. =(
     
  10. Mar 10, 2010 #9
    Ok, I think the only mistake you made was this conversion. It's acceleration, therefore has a unit of either [tex]m/s^{2}[/tex] or [tex]km/h^{2}[/tex], notice the square in the bottom. I tried converting it and got [tex]1m/s^{2}=12960km/h^{2}[/tex]
     
  11. Mar 10, 2010 #10
    How did you get that?
     
  12. Mar 10, 2010 #11
    Now I just confused 2 questions together. Forget this whole thing, it's a flop. =(
    I'll ask my teacher.
     
  13. Mar 10, 2010 #12
    Ok, this is what I did.
    [tex]1km/h^{2}=1000m/(3600s)^{2}[/tex]
    Try and see what you get.
     
  14. Mar 10, 2010 #13
    Oh, so I have to square the seconds in the speed part?
     
  15. Mar 10, 2010 #14
    of course!
     
  16. Mar 10, 2010 #15
    Now I got -0.1296..
    Kinda on the right track.

    I did -0.006/(0.36^2)
     
  17. Mar 10, 2010 #16
    nope, still wrong. 1/3600 is not equal to 0.36.
     
  18. Mar 10, 2010 #17
    Is this considered partially right?
    0.006km/3600 seconds?
     
  19. Mar 10, 2010 #18
    You really need to work on unit conversion. Let me show you:
    [tex]1km/h^{2}=1000m/(3600s)^{2}[/tex] 1km is replaced by 1000m, and 1h is replace by 3600s.
    [tex]1km/h^{2}=1000m/(3600^{2}s^{2})[/tex] notice that both the 3600 and s are squared.

    [tex]1km/h^{2}=\frac{1000}{3600^{2}}\frac{m}{s^{2}}[/tex] separate the unit from the numbers.

    [tex]1km/h^{2}=\frac{1}{12960}\frac{m}{s^{2}}[/tex]

    [tex]12960km/h^{2}=1\frac{m}{s^{2}}[/tex]

    Right?
     
    Last edited: Mar 10, 2010
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