A particle moves along a straight line such that its position is given by x(t) = 2t^3 - 9t^2 + 12*t + 5, where t is in seconds. At what times is the particle momentarily at rest?
Answer and explanation
Correct answer: A. t = 1 s and t = 2 s
- Step 1: Velocity is the derivative of position: v(t) = dx/dt = 6t^2 - 18t + 12.
- Step 2: The particle is at rest when v(t) = 0: 6(t^2 - 3t + 2) = 0.
- Step 3: Factor: 6*(t - 1)(t - 2) = 0.
- Step 4: Thus, v(t) = 0 at t = 1 s and t = 2 s.
Answer
t = 1 s and t = 2 s.
Why not the others: t = 0.5 s gives v(0.5) = 4.5 m/s != 0; t = 1.5 s is the time of minimum velocity (inflection point of position); t = 2 s and 3 s has v(3) = 12 m/s.
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