(A) Momentum is given by the equation `p = mv` , so the initial momentum is `(950 kg)(25 m/s) = 23,750 kg*m/s` . Since we are given two significant figures, we should round to two significant figures for our answers, so the final answer is 2.4 *10^4 kg*m/s.

(B) The impulse is the change in momentum, so we need to find the final momentum, again by `p = mv` . The final momentum is `(950 kg)(17 m/s) = 16,150 kg*m/s` , which in two significant figures is 1.6 *10^4 kg*m/s.

(C) The direction of the force (applied by the friction of the brakes) is backwards, resisting the motion of the car. Its magnitude is the impulse divided by the time, `F = {Delta p}/{Delta t} = (16,150 kg*m/s)/(3.2 s) = 5046.875 N` , which in two significant figures is 5.0*10^3 N.

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