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Movement - Deceleration

  1. Jan 9, 2005 #1


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    This, to most of you, is probably easy to work out.
    I have the following question:

    A car has a mass of 900kg and is travelling at 35m/s on a motorway.
    The driver sees that a queue has developed and applied the brakes to give the car a decelerating force of 2000N until it comes to rest. What distance does the car travel while it is decelerating?

    Which formula should I use for this?

    Thanks for your help :smile:
  2. jcsd
  3. Jan 9, 2005 #2
    [tex]x - x_0 = v_0.t -a \frac{t^2}{2}[/tex]

    the LHS denotes the travalled distance and v_0 denotes the initial velocity which is given. Also the acceleration a is given as [tex]a = \frac{F}{m}[/tex]

    You know F and you know the mass

    In order to find the time just apply
    [tex]v = 0 =v_0 -at[/tex]
    Last edited: Jan 9, 2005
  4. Jan 9, 2005 #3


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    METHOD 1.

    If your force is constant,use Galileo Galilei's formula
    [tex] v_{f}^{2}=v_{i}^{2}+2ax [/tex](1)
    [tex] a=\frac{F}{m} [/tex](2)
    (pay attentian,it is negative,it's a decelerating force).
    [tex] x=-\frac{mv_{i}^{2}}{2F}[/tex](3)

    METHOD 2.

    Use the thoerem of variation if KE.It will lead your to (3) in no time.


    PS.These are two alternative option to the more complicated method proposed by my fellow Marlon (finding the accleration,the time it takes to stop,squaring the time,making multiplications and additions,chosing an origin for time and 'x'...)
  5. Jan 9, 2005 #4
    As marlon said, [tex] a = \frac {F}{m}[/tex]

    You can use that equation of motion...
    [tex] v^2 = u^2 + 2 a s[/tex] where a being [tex] - \frac {F}{m}[/tex] because it is decelerating, u and v being the initial (o m/s) and final (35 m/s) velocities respectively. Replacing the values in the above equation, we have...

    [tex] s = \frac {u^2 m} {2 F}[/tex]
  6. Jan 9, 2005 #5
    chose an origin for time and x ??? hahahahaha
    that is so sad man...just take the origin to be ZERO...no nitpicking please...

    And besides, i think this way will yield the exact formula that you used wouldn't you say...Don't you have to make those terrible additions and multiplications ??? :rofl: :rofl: :rofl: dexter you make me laugh man... :rolleyes:

    marlon eeuuuuuuuuuh
  7. Jan 9, 2005 #6
    Dexter. I think you need to remove the negative sign from your formula. What do you say?
  8. Jan 9, 2005 #7


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    As i told u on the other thread,over the last couple of days,u've been sooooooooo funny... :rofl:
    [tex]\Delta KE (1\rightarrow 2)=W_{1\rightarrow 2} [/tex]

  9. Jan 9, 2005 #8


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    Nope.The force is negative.The "x" will come out positive.


    PS.The Ox axis has the sense that coincides with the sense of the velocity and is opposite to the one of acceleration and force.
  10. Jan 9, 2005 #9


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    Thanks for your help :smile:
  11. Jan 9, 2005 #10
    Alright Dex, I got it. I thought you were about to use a positive value for the force.

  12. Jan 9, 2005 #11

    Thanks man...i wish i could say the same about you :wink:

  13. Jan 9, 2005 #12

    What a strange name ??? How do you justify this formula whithout using Newton's modell ?

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