physics

Hello!
I'm studying for physics and I am stuck on this question (I have the answer):
Two identical cars are travelling on a horizontal road, one of them is going at twice the speed of the other one. Both break using the same force. The slower car breaks completely after 20m. At what distance did the faster car come to a full stop? The answer was 80m. I just don't know how to arrive to it. Can anyone help? Thank you!

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asked by castiel
  1. Newton's second law:
    F=ma
    if F and m are both identical, the a would be the same, where a=deceleration in m/s².

    Using the kinematic equations for the first car,
    vf²=vi²+2aΔx
    vf=final velocity=0
    vi=initial velocity
    Δx=stopping distance
    and solve for Δx.
    Δx = -vi²/2a = 20 m
    where a<0 for deceleration.

    For the second car, vi is doubled, so
    -(2vi&sp2;)/2a
    =-4vi²/2a
    =80 m

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    posted by MathMate
  2. I saw this earlier and I tried using kinematic equations, but I could't reason 80m as the answer. When you plug in a value for a that is less then or greater then 0, you do not calculate 80m, at least I didn't. So, I am not sure what MathMate did, and I do not think that it is correct.

    Use the following relationship, which is based on the conservation of mechanical energy:

    Work=Kinetic Energy

    F*d=1/2mv^2

    Where

    F=force
    d=displacement
    v=velocity
    and
    m=mass

    The cars are identical, so F is the same for both cars. Solve for !@#$%^&
    F*d=1/2mv^2

    F=(1/2mv^2)/d

    For the first car, the car that stops a distance of 20m, the equation becomes the following:

    F*20m=1/2mv^2

    Solving for !@#$%^&
    F=(1/2m^2)/20m

    *** The question states that the cars experience the same force and are identical.

    Since F for both are cars are the same, set the equations equal to each other.

    (1/2m^2)/d=(1/2m^2)/20m

    Masses are equal for both cars and cancel out since the cars are identical.

    (1/2v^2)/d=(1/2v^2)/20m

    Further simplification gives the following:

    v^2/d=v^2/20m

    *** Remember, the second car has a velocity twice that of the first car

    Let v1= velocity of the first car and v2= velocity of the second car.

    v2=2v1, so the equation becomes the following:

    [(2v)^2/]d=v^2/20m

    4v^2/d=v^2/20m

    Taking the inverse of the equation, the equation becomes the following:

    d/4v^2=20m/v^2

    Solving for d:

    d=20m*(4v^2/v^2)

    d=20m*4

    d=80m


    I hope this explanation wasn't too late, and best and good luck.

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    posted by Devron
  3. thank you so much!

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    posted by castiel

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