A car moving initially with speed v₀ slows down with an acceleration of magnitude a and comes to a full stop after traveling a distance d. What was the speed of the car when it had traveled half that distance, d/2? A) v₀/2 B) v₀/4 C) v₀/ $\sqrt { 2 }$ D) v₀/8

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Let t be the time taken for the car to come to a full stop after travel d. Then, we have: d = v₀t + (1/2)at^2 (1) At the halfway point, the car has traveled a distance of d/2, so: d/2 = (1/2)v₀t + (1/8)at^2 (2) From equation (1), we can solve for t: t = (v₀ - 0)/a = v₀/a Substituting into equation (2), we get: d/2 = (1/2)v₀(v₀/a) + (1/8)a(v₀/a)^2 Simplifying, we get: d/2 = (1/2)v₀t + (1/8)a(v₀)^2/a^2 d/2 = (1/2)v₀t + (1/8)(v₀)^2/a Solving for v₀, we get: v₀ = (d/2 - (1/2)v₀t) * (8/a) Substituting t = v₀/a, we get: v₀ = (d/2 - (1/2)v₀ * v₀/a) * (8/a) Multiplying both sides by a and rearranging, we get: (v₀)^2 - (4ad) = 0 Therefore, v₀ = +/-sqrt(4ad). Since the car is slowing down, we take the negative root. Plugging in the values, we get: v₀ = -sqrt(4ad) At the halfway point, the car has traveled a distance of d/2, so we can plug that into equation (1) to solve for the speed at that point: (d/2) = v₀(t/2) + (1/2)at^2/4 (d/2) = v₀^2/2a + (1/8)at^2 Substituting t = v₀/a, we get: (d/2) = v₀^2/2a + (1/8)v₀^2 Simplifying, we get: v₀^2 = (d/2) * (16a/9) Taking the square root, we get: v₀ = sqrt((d/2) * (16a/9)) Simplifying, we get: v₀ = v₀/ $\sqrt { 2 }$ Therefore, the answer is C.

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