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Velocity, acceleration and distance

  1. Apr 18, 2009 #1
    1. The problem statement, all variables and given/known data
    An object is pushed from rest across a sheet of ice, accelerating at 5.0 m/s2 over a distance of 80 cm. The object then slides at a constant speed for 4.0 s until it reaches the rough section that causes the object to stop in 2.5 s.

    a) What is the speed of the object when it reaches the rough section?
    b) At what rate does it slow down once it reaches the rough section?
    c) What is the total distance that the object slides?



    Any help will be appreciated.




    2. Relevant equations




    3. The attempt at a solution

    I think:
    a) find v
    b) find a
    c) find d(distance)

    I have some answers but I don't think its correct because of the rough section part of the question.
     
  2. jcsd
  3. Apr 19, 2009 #2

    LowlyPion

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    OK. What do you get for the speed when it reaches the rough area?
     
  4. Apr 19, 2009 #3
    I got

    a) 2.8 m/s
    b) -1.12 m's2
    c) 15.5 m
     
  5. Apr 19, 2009 #4

    LowlyPion

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    a) looks correct. So does b) I get a slightly different rounding.
     
  6. Apr 19, 2009 #5
    b) a = (Vf - Vi)/t
    a = (0 - 2.8 m/s)/2.5 s
    a = - 1.12 m/s^2 answer
     
  7. Apr 19, 2009 #6

    LowlyPion

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    It's not a big difference, but I used the √8 for v.

    Now c) then is 4 * a) plus the calculation of the distance derived from slowing down over the rough. So what is the deceleration distance?
     
  8. Apr 19, 2009 #7
    This is what I did to get c)

    total distance = distance when it was being pushed + distance when
    it slides at constant speed for 4.0 s + distance during which it
    decelerates

    Dt = D1 + D2 + D3
    Dt = [(2.8 m/s)^2 - 0^2]/[(2)(5.0 m/s^2)] + (2.8 m/s)(4.0 s)
    + [0^2 - (2.8 m/s)^2]/[(2)(- 1.12 m/s^2)]
    Dt = 15.5 m answer
     
  9. Apr 19, 2009 #8

    LowlyPion

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    The question asks for how far it slides. I wouldn't include the .8m it was pushed.

    That yields √8 * 4 = 11.314
    8/(2*1.13) = 3.54
    Total 14.85 m.
     
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