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Negative value when calculating the distance

  1. Oct 23, 2011 #1
    A 1000 kg car travelling rightward at 25 m/s slams on the breaks and skids to a stop (with locked wheels). Find the distance required to stop it. Coefficient of friction between the tires and the road is 0.95. Let g=9.8 m/s^2

    Solution:
    2aΔx=vf2- vo2
    a=µg
    2 µg Δx=vf2- vo2
    vf=0
    2 µg Δx=- vo2
    Δx=- vo2/2 µg
    Δx=-252/(2×0.95×9.8)
    Δx=-33.6 m

    The answer became -33.6 m. Why is the value negative? The correct answer is 33.6. Thank you in advance!
     
  2. jcsd
  3. Oct 23, 2011 #2

    ehild

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    The acceleration is negative, a=-µg.

    ehild
     
  4. Oct 23, 2011 #3
    If I have understood this correctly, the solution should be this:

    2aΔx=vf2- vo2
    a=-µg
    2 -µg Δx=vf2- vo2
    vf=0
    2 -µg Δx=- vo2
    Δx=- vo2/2 -µg
    Δx=-252/(2×-0.95×9.8)
    Δx=33.6 m
     
  5. Oct 23, 2011 #4

    ehild

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    To denote power, use "^". Use parentheses when multiplying with a negative number. 2 -µg Δx means subtraction. Correctly: 2(-µg)Δx or -2µg Δx. And 25^2=625.

    Written correctly your derivation, it looks :

    2aΔx=vf^2- vo^2
    a=-µg
    2 (-µg) Δx=vf^2- vo^2
    vf=0
    2 (-µg)Δx=- vo^2
    Δx=- vo^2/(-2µg)
    Δx=-625/(2×(-0.95)×9.8)
    Δx=33.6 m

    ehild
     
  6. Oct 23, 2011 #5
    Thank you very much ehild! I understand now.
     
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