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Equations of Motion Question

  1. Oct 28, 2009 #1
    A car travelling at 40km/h has it's speed reduced to 10km/h in 2.0s. Assuming that the same deceleration would be in effect, find how far the car will travel in coming to rest from a speed of 60km/h?
     
  2. jcsd
  3. Oct 28, 2009 #2

    hage567

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    What are you thoughts on how to start this problem?
     
  4. Oct 28, 2009 #3
    I'm thinking about calculating the acceleration using the givens from the first part, to calculate the second part.

    so using the equation,
    v2^2 = v1^2 + 2(a)(t)

    But what has gotten me really puzzled is the seconds, the hours, and the kilometers.
     
  5. Oct 28, 2009 #4

    hage567

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    That's the right approach, but that equation is not quite right. That t should be a d (as in distance). Check for another equation.

    You will need to convert your quantities so they are all in the same units. I usually work in meters and seconds in these questions. I would start with that before you do anything else.
     
  6. Oct 28, 2009 #5
    Okay, so after converting all the units to m/s..

    first part:
    v1 = 40km/h = 11.11m/s
    v2 = 10km/h = 2.8m/s
    t=2.0s
    a = v/t = (2.8m/s - 11.11m/s) / 2.0s = -4.155m/s^2

    second part:
    v1 = 60km/h = 16.67m/s
    v2 = 0km/h = 0m/s
    a = -4.155m/s^2
    d = ?

    v2^2 = v1^2 + 2(a)(d)
    0 = (16.67)^2 + 2(-4.155)(d)
    0 = 277.9 + (-831)(d)
    d = 831 - 277.9
    d = 553.1m/s

    However, the correct answer to this question is 0.033km..
     
  7. Oct 28, 2009 #6

    hage567

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    That -831 is wrong, you missed the decimal place!

    Also, you didn't solve for d correctly. There should be division involved. Try again.
     
  8. Oct 28, 2009 #7
    v2^2 = v1^2 + 2(a)(d)
    0 = (16.67)^2 + 2(-4.155)(d)
    0 = 277.9 + (-8.31)(d)
    -277.9 / -8.31 = (-8.31)(d) / (-8.31)
    d = 33.44m

    so, 33.44/1000 = 0.033km!

    thank you so much :smile:
     
  9. Oct 28, 2009 #8

    hage567

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    You're welcome!
     
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