Question

Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x'(t) is its velocity, and x''(t) is its acceleration. x(t) = t^3 − 12t^2 + 21t − 9, 0 ≤ t ≤ 10.

(c) Find the velocity of the particle when the acceleration is 0.

Answer 1

To find the velocity when the acceleration is 0, take the derivative of the position function and set it equal to 0. Solve for t to find the times when the acceleration is 0. Substitute these times back into the velocity function to find the velocity.

Step-by-step explanation:

To find the velocity of the particle when the acceleration is 0, we first find the acceleration function by taking the derivative of the position function twice. The derivative of x(t) = t^3 - 12t^2 + 21t - 9 is a(t) = 6t - 24t + 21. To find when the acceleration is 0, we set a(t) = 0 and solve for t: 6t - 24t + 21 = 0.

Simplifying this equation gives us t = 7/6 or t = 7/2. Substituting these values back into the velocity function x'(t) = 3t^2 - 24t + 21, we find that the velocity when the acceleration is 0 is 0 at t = 7/6 and again at t = 7/2.