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Finding displacement through velocity and deceleration

  1. Feb 17, 2016 #1
    1. The problem statement, all variables and given/known data
    a car with an initial velocity of 10.0 m/s (W) decelerates at -2.00 m/s (s to the power of 2) (W) reaching a final velocity of 5.00 m/s (W). What is the displacement of the car?

    2. Relevant equations
    Δdisplacement = (final velocity to the power of 2 - initial velocity to the power of 2)/ 2 times acceleration

    3. The attempt at a solution
    I didn't really get far because I didn't know how to deal with the units and the minus/plus signs
     
  2. jcsd
  3. Feb 18, 2016 #2

    cnh1995

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    I believe it should rather be 'accelerates at -2m/s2'. Negative acceleration is deceleration and negative deceleration is acceleration..
     
  4. Feb 18, 2016 #3
    true, I didnt know how to add the subscript on top. But can you help me answer this?
     
  5. Feb 18, 2016 #4

    cnh1995

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    Your relevent equation is correct. Put the known values in that equation and solve for displacement.
    s=(v2-u2)/2a
     
  6. Feb 18, 2016 #5
    I dont know how to. The anwer im getting is -18.75, and i dont think that's right.
     
  7. Feb 18, 2016 #6

    cnh1995

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    It is negative because you used a=2m/s2
    Since the car is decelerating, you should use a=-2m/s2..
    Then you'll get it as 18.75m.
     
  8. Feb 18, 2016 #7
    This is what i did-
    (5.00 m/s [W])2 - (10.0 m/s [W])2 / 2 * -(-2.00 m/s2
    (-5.00 m/s)2 - (-10.0 m/s)2 / 4.00 m/s2
    After this, what happens to the units? How do the s2 cancel each other out?
     
  9. Feb 18, 2016 #8
    Do the units become m2/s2 divided by m/s2?
     
  10. Feb 18, 2016 #9

    cnh1995

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    Yes. And only the unit of displacement remains which is meter..Every equation has to be dimensionally correct.
     
  11. Feb 18, 2016 #10
    So s2 divided by s2 cancels each other out and does not equal s4? Because acceleration is m/s divided by s
     
  12. Feb 18, 2016 #11

    SteamKing

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    Why did you put double negative signs on the acceleration in the denominator?

    You should be able to work out the resulting units using standard algebra.

    If you square velocity in m/s, what do you get?

    If you divide these units for velocity squared by the units for acceleration, what's left?
     
  13. Feb 18, 2016 #12

    cnh1995

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    Yes. You can see s2 cancel each other and m2/m becomes m.
     
  14. Feb 18, 2016 #13
    I put double negatives on the denominator because the question says is "decelerates at -2.00 m/s2 [W] or -(-2.00 m/s2)"
     
  15. Feb 18, 2016 #14
    Is this wrong? If so, how would I write this out?
     
  16. Feb 18, 2016 #15

    SteamKing

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    The question is poorly worded. One negative sign should suffice, since 'decelerate' means to reduce speed or velocity.
     
  17. Feb 18, 2016 #16

    cnh1995

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    The initial and final velocities are 10m/s and 5m/s respectively. This clearly tells you that the car is 'decelerating'. So, you should write either 'a car is decelerating at 2m/s2' or ' a car is accelerating at -2m/s2'. That's what I meant in #2.
    You should write the equation in the following form.
    Now substitute a=-2 here, since the car is decelerating.
     
  18. Feb 18, 2016 #17
    Thank you so much for your help!
     
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