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Kinematics- police driver acceleration problem

  1. Aug 16, 2012 #1
    In an accident investigation police measured the skidmarks left my vehicle. they measure the skidmarks to be 150 m from the point where the driver applied the brakes to where the car came to rest. In their investigation the police Sumat the driver was traveling at this bead limit of 80 km/h (22m/s) when the brakes were applied. Calculate the acceleration of the vehicle.

    ok so i know acceleration is

    d= delta v
    ----------
    delta t

    = vf - vi
    -------
    delta t

    but because this question has no initial point. What do I use? Do I just started off as zero OR do I have to solve for something? can someone please help me. thank you
     
  2. jcsd
  3. Aug 16, 2012 #2

    CAF123

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    Consider the kinematic relation [tex] v_{fx}^2 = v_{ox}^2 + 2as,[/tex] where [itex] v_{fx} = [/itex] final velocity of car after skid, [itex] v_{ox} = [/itex] velocity of car before skid and[itex]\,[/itex] [itex] s = [/itex] displacement.

    Think about what this equation is saying: From the stage of the journey where the car was travelling at 80km/h to the stage of the journey where the car is at rest, it has covered a distance of 150m, and so we can determine it's acceleration using the above relation.
     
  4. Aug 16, 2012 #3
    ok so my s is 150m and my initial veloticy is 80km but i need to put that into m/s.. how do i find the final velocity?
     
  5. Aug 16, 2012 #4

    CAF123

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    The final velocity is given to you in the question. The words 'at rest' mean the car has halted to zero velocity.
     
  6. Aug 16, 2012 #5
    ok so its just 0... so i just sub everything into the equaiton?
     
  7. Aug 16, 2012 #6

    CAF123

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    Yes. Notice the negative sign for acceleration. What does this mean?
     
  8. Aug 16, 2012 #7
    because it is slowing down thats why it is negative
     
  9. Aug 16, 2012 #8

    CAF123

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    Yes, but to be a little bit more precise, if the car is defined to be moving in positive x direction and slowing down, then the acceleration vector would point in the negative x direction. Thus we get a 'negative acceleration' ( which is more precise than the word 'deceleration' used in everyday life)
     
    Last edited: Aug 16, 2012
  10. Aug 16, 2012 #9
    ok so 80km/h is 22m/s

    so..

    0=22^2+2a(150)
    0= 484+a300
    -484=a300

    -484
    ------ = a
    300

    -1.61 = a is it still in m/s?
     
  11. Aug 16, 2012 #10
    or should i have stuck with 80km... and i would of got -21.3 km/h (i think thats the units not ti sure)
     
  12. Aug 16, 2012 #11

    CAF123

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    It is always best to rearrange the equation for the given quantity required. (This does help in figuring out the units).

    So the kinematic relation rearranged for [itex] a [/itex] yields [tex] a = \frac {v_{fx}^2 - v_{ox}^2}{2s}, [/tex]
    If we replace each physical quantity with its units we end up with [tex] a= \frac{m^2/s^2}{m} [/tex] from which you see one [itex] m [/itex] cancels top and bottom to give m/s2 as the units.

    However, having said that, with these sorts of quantities that we use 'all the time' in physics, they become second nature after a while. So this 'dimensional analysis', as it is called, is not required all the time, only when dealing with unfamiliar expressions.
     
  13. Aug 16, 2012 #12

    CAF123

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    The definition of acceleration is the change in velocity per unit time. So this -1.61 m/s2 means the car slows down at a rate of 1.61 m/s every second.
     
    Last edited: Aug 16, 2012
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