Physics Part A HELP!!

4.) In coming to a stop, a car leaves skid marks 80m long on the highway. Assuming a DEceleration of 7.00 m/s^2 (m/s/s), estimate the speed of the car just before breaking.

6.) A car traveling @ 90km/h strikes a tree. The front end of the car compresses and the driver comes to rest after traveling 0.80m. What was the average acceleration of the driver during the collision? Express answer in terms of ("g's") where 1.00 g = 9.80 m/s^2 OR (m/s/s)

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  1. I found an example like #6...

    A car traveling at 95m/h strikes a tree. The front end of the car compresses and the driver comes to rest after traveling 0.80m. What was the average acceleration of the driver during the collision? Express the answer in terms of "g's," where 1.00g = 9.80m/s2.

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    Here's the WORK:

    v2 = vo2 + 2aav(x2 - x1);

    0 = [(95km/h)(3.6ks/h)]2 + 2aav(0.80m), which gives aav = -4.4 x 102m/s2.

    The number of g's is
    | aav | = (4.4 x 102m/s2)/[(9.80m/s2)/g] = 44g.

    **I DON'T GET WHERE THE (3.6 ks/h) CAME FROM. HOW DID THE PERSON GET 3.6?? **

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  2. 4) x=(1/2)a*t^2
    t=sqrt(2x/a)
    plug t into --> v=a*t

    6) 90km/h = 25m/s
    v=at
    a=v/t --> a=25/t
    plug in a and solve --> x=x+vt+(1/2)at^2
    0.8 = 25t - (1/2)(25/t)t^2
    That gives you t which you can then plug back into a=v/t to get acceleration. You divide that by 9.8m/s^2 to give you g's.

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  3. LOST! I'm confused putting the number in!!
    Test tmrw!!!

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  4. Ok, starting from 0.8=25t-(1/2)(25/t)t^2

    0.8 = 25t - (25t^2/2t)

    0.8 = 25t - 12.5t

    0.8 = 12.5t

    t = 0.8/12.5

    Plug that back into our first equation:

    velocity = acceleration * time

    a = v / t

    a = 25m/s / (0.8/12.5)

    a / 9.8 = g force

    Plug in values and and prosper.

    You should get a g force of approx 39.86

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  5. Shouldn't my answer be negative 39.86?

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