A particle moves in a straight line with acceleration a(t) = 6t - 12 for t >= 0. If initial velocity v(0) = 9 and initial position s(0) = 2, what is the position function s(t)?
Answer and explanation
Correct answer: D. s(t) = t^3 - 6t^2 + 9t + 2
- Step 1: Step 1: Velocity v(t) = integral(a(t) dt) = 3t^2 - 12t + C_1. Using v(0) = 9 gives C_1 = 9 => v(t) = 3t^2 - 12t + 9. Step 2. Step 2: Position s(t) = integral(v(t) dt) = t^3 - 6t^2 + 9t + C_2. Using s(0) = 2 gives C_2 = 2 => s(t) = t^3 - 6t^2 + 9t + 2. Answer: s(t) = t^3 - 6t^2 + 9t + 2. Why not the others: Integrating acceleration once gives velocity; a second integration gives position.
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