If velocity is given by v(t) = 3t^2 - 12t + 9 for t >= 0, at what times is the particle at instantaneous rest?
Answer and explanation
Correct answer: C. t = 1 and t = 3
- Step 1: Particle is at rest when v(t) = 0. Step 2. Solve: 3t^2 - 12t + 9 = 0 => 3(t^2 - 4t + 3) = 0. Step 3. Factor: 3(t - 1)(t - 3) = 0. Step 4. Roots are t = 1 and t = 3. Answer: t = 1 and t = 3. Why not the others: t = 2 is where acceleration is zero; t = 0 gives v = 9 != 0.
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