A particle moves according to the law of motion s(t) = t^3 - 8t^2 + 4t, t >= 0, where t is measured in seconds and s in feet. a.) Find the velocity at time t. Answer: b.) What is the velocity after 3 seconds? Answer: -8 c.) When is the particle at rest? Enter your answer as a comma separated list. Enter None if the particle is never at rest. At t1 = 2-sqrt3 and t2 = 2+sqrt3 with t1 < t2. d.) When is the particle moving in the positive direction? When 0 <= t < 2-sqrt3 and t > 2+sqrt3
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s(t) = -8t^2 + 4t + 12 Taking the derivative: v(t) = ds/dt = -16t + 4 So the velocity function is v(t) = -16t + 4. b) Show more…
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