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?

I know you use the equation d = Vit +1/2 at^2 but to find the distance would you just calculate d = vit only.

v = Vo + a t
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

Vo = 110 m/s

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  1. Ok i think i understand. I'm pretty sure you have to use the full equation d= vit +1/2 at^2

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  2. a) The average velocity when decelerating will be Vo/2 = 27.5 m/s. Multiply that by the time required to stop, Vo/a, to get the stopping distance X.

    X = (Vo/a) * (Vo/2)
    X = Vo^2/(2 a)= (55)^2/(22) = 137.5 m

    Same answer you got, but I used a shortcut

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  3. HAPPY NEW YEAR! Ok, thanks a lot.

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