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Acceleration of skidding car

  1. Jan 17, 2014 #1
    1. The problem statement, all variables and given/known data

    Attempting to stop on a slippery road, a car moving at 80 km/h skids across the road at a 30 degree angle to its initial motion, coming to a stop in 3.9 s.

    Determine the average acceleration in m/s^2, using a coordinate system with the x axis in the direction of the car's original motion and the y axis toward the side of the road to which the car skids.


    3. The attempt at a solution

    I did a whole chunk of workings but they're aren't working out to give me a sensible answer.
    Could someone give me a leg up?

    0ms^-1 = 22.2ms^-1 + acos30°(3.9s)
    a = -6.57ms^-2

    vx = 22.2ms^-1 + (-6.57ms^-1 cos30)(3.9s)
    vy = (-6.57ms^-1 sin30)(3.9s)

    I square root the square of vx and vy to get the resultant, then, divide |v| by t = 3.9s but it's not yield -5.7ms^-1.

    What is wrong here?
     
    Last edited: Jan 17, 2014
  2. jcsd
  3. Jan 17, 2014 #2
    Why do you even need a coordinate system here? Is that a requirement for this problem?

    If so, none of your equations makes sense for me. If the initial velocity is at 30 degrees to the x axis, what are its x and y components?

    Then, the final velocity must have zero x and y components. Determine the components of acceleration required for that.
     
  4. Jan 17, 2014 #3
    Isn't it the acceleration magnitude that is at 30 degrees to the x-axis?

    ax = |a|cos30°
    ay = |a|sin30°
     
  5. Jan 17, 2014 #4
    "skids across the road at a 30 degree angle to its initial motion" is a condition on velocity.
     
  6. Jan 17, 2014 #5
    If so, then I've obtained the answer.

    |a→| = Δv/Δt = [SQRT(22.2ms^-1 cos 30°)^2 + (22.2ms^-1 sin 30°)^2]/3.9s

    |a→| = 5.7ms^-2
    a→ = -5.7ms^-2
     
  7. Jan 17, 2014 #6
    Look what you did. (1) You found the magnitude of velocity: 22.2 m/s. (2) Then you decomposed it into the x and y components. (3) Then you found the magnitude of velocity from the components. (4) Then you divided it by time.

    Were steps (2) and (3) really required?
     
  8. Jan 17, 2014 #7
    It was not-22.2ms^-1/3.9s= 5.69ms^-1 - but doing so allows me to better appreciate what is going on.
     
  9. Jan 17, 2014 #8
    Then I suggest something else.

    Once you know the components of velocity, find what acceleration is required for each component to become zero in 3.9 s. This is a rephrasing of my post #2.
     
  10. Jan 17, 2014 #9
    x-component of velocity = 19.22ms^-1
    y-component of velocity = 11.1ms^-1

    vxf = vxi +Δv = 22.2ms^-1 + 19.22ms^-1 = 41.42ms^-1
    (my reasoning is based on the premise the initial velocity = 22.2ms^-1 but at the point at which p(0,0), the car skids at 30° to the x-axis with a velocity of magnitude 22.2ms^-1.

    vyf = vyi + Δv = 0ms^-1 + 11.1ms^-1

    |a→|x = Δvx/Δt = [vxf-vxi]/3.9s = -41.42ms^-1 / 3.9s = -10.6205ms^-2
    |a→|y = Δvy/Δt = [vxf-vxi]/3.9s = -11.1ms^-1 / 3.9s = -2.846ms^-2
     
  11. Jan 17, 2014 #10
    Again, the initial velocity is 80 km/h at 30 degree. You cannot count i t twice, it makes no sense.
     
  12. Jan 17, 2014 #11
    I did contemplate about this quandary-and I'm sure you are right in your statement-but I was thinking along the line of 80kmh^-1 + velocity of 80kmh^-1 at 30 degree at the instantaneous moment when it skid to give vi....
     
  13. Jan 17, 2014 #12
    otherwise,

    (-19.22ms^-1 , -11.1ms^-1)/3.9s = (-4.9ms^-2 , -2.846ms^-2)
     
  14. Jan 17, 2014 #13
    There is only one velocity at any given moment of time. If it is given to be 80 km/h at some angle, then its x component is 80 km/h times the cosine, and the y component is 80 km/h times the sine. You cannot say, oh, it was 80 km/h "straight" just a moment ago, let's add that to what it is now - because then you can just say, wait, it was some other value just one and a half moment ago, so let's add that as well - where do you stop in that sort of reasoning? One velocity at any one time, plain and simple.
     
  15. Jan 17, 2014 #14
    What magnitude of acceleration does that yield? Is that consistent with the other method?
     
  16. Jan 17, 2014 #15
    Yes, this returns me the value 5.7ms^-2-consistent with the average acceleration.
     
    Last edited: Jan 17, 2014
  17. Jan 17, 2014 #16
    I anticipated this and was aware of the mathematical and logical implication..
     
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