Physics

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A race car can be slowed with a constant acceleration of -11 m/s^2

a. If the car is going 55 m/s, how many meters will it travel before it stops?

b. How many meters will it take to stop a car going twice as fast?

  • Physics -

    v = Vo + a t
    so
    v = 55 - 11 t
    when is v = 0 ?
    0 = 55 - 11 t
    so t = 5 seconds to stop
    d = Vo t + (1/2) a t^2
    d = 55 (5) - (1/2) 11 (25)
    = 275 - 137.5
    =137.5 meters
    Now do that again for Vo = 110 m/s

  • Physics -

    Ok, thanks but i don't understand b. I know you use the equation d = Vit +1/2 at^2 but to find the distance would you just calculate d = vit only.

  • Physics -

    No, do v = Vo + a t again with
    0 = 110 - 11 t

    then use that t in
    d = 110 t - (1/2)11 t^2

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