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  1. #1
    Member classicstrings's Avatar
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    Vectors`

    Ok .. here it is...

    The velocity of a train is reduced from 30m/s to 10m/s over a distance of 200m. Under this constant deceleration the train travels a further distance before coming to rest. What is this distance?
    Last edited by classicstrings; October 29th 2006 at 02:01 AM.
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  2. #2
    Member classicstrings's Avatar
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    Posted the question now.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by classicstrings View Post
    Ok .. here it is...

    The velocity of a train is reduced from 30m/s to 10m/s over a distance of 200m. Under this constant deceleration the train travels a further distance before coming to rest. What is this distance?
    For the record I am assigning a positive direction in the direction of the initial velocity of the train. I am temporarily assigning an origin to be at the point where the train has a speed of 30 m/s.

    To find the acceleration of the train:
    v^2 = v_0^2 + 2a(x - x_0)

    10^2 = 30^2 + 2 \cdot a \cdot 200

    [tex]a = \frac{100 - 900}{2 \cdot 200}

    a = -2 \, m/s^2

    (The negative sign merely indicates that the acceleration is in the opposite direction to the initial velocity, meaning that it is a decleration in this case.)

    Now I will reset the origin to the point where the speed is 10 m/s. We wish to find out how far the train travels from this point until it comes to rest. We may use the same equation:

    v^2 = v_0^2 + 2a(x - x_0)

    0^2 = 10^2 + 2(-2)(x - 0)

    x = \frac{0 - 100}{2 \cdot -4}

    x = 25 \, m

    -Dan
    Last edited by topsquark; October 29th 2006 at 02:55 AM.
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    Member classicstrings's Avatar
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    Hello Topsquark thanks for the answer, i'm still looking at it, but my solutions (not that I know why it is 25) says it is meant to be 25metres. Thanks!

    EDIT: Ok I see your calculation error, -800 = 400a, a = -2m/s^2
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  5. #5
    Member classicstrings's Avatar
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    I get it now, thanks! It works out, after the a = -2m/s^2.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by classicstrings View Post
    I get it now, thanks! It works out, after the a = -2m/s^2.
    Whoops! Sorry about that. (Go figure, the one calculation I trust myself to do in my head...) I fixed it in my original post.

    -Dan
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    Member classicstrings's Avatar
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    Quote Originally Posted by topsquark View Post
    Whoops! Sorry about that. (Go figure, the one calculation I trust myself to do in my head...) I fixed it in my original post.

    -Dan
    Thanks a lot! Would you be able to give a quick look to the other one i just posted please?
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