# Bill uniformly slows down from 15.0 m/s to 0 m/s in 2.50 s How many m b fore a stop sign must apply breaks. just a practice prob?

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- Anonymous1 month ago
distance d = ΔV*t/2 = (15-0)*2.5 / 2 = 7.5*2.5 = 18.75 m

or

distance d = (0-V^2) / (2*(0-V)/t) = -(15^2) / (-30/2.5) = -225*-2.5/30 = 18.75 m

or

distance d = V*t+(0-V)/2t*t^2 = V*t-15/5*2.5^2 = 15*2.5-3*6.25 = 18.75 m

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