Question

A particle moves along the x axis so that at any time its position is given by x(t)=t^3-3t^2-9t+1. For what values of t is the particle at rest?

Answer 1

The particle described by the position function x(t) is at rest at t = 3 seconds.

We find this by differentiating the position function to get the velocity function and then setting it equal to zero to solve for t.

Step-by-step explanation:

To find out when the particle described by the position function x(t) = t^3 - 3t^2 - 9t + 1 is at rest, we need to determine when the velocity of the particle is zero.

The velocity function is the derivative of the position function with respect to time.

V(t) = dx/dt = 3t^2 - 6t - 9.

We set the velocity function equal to zero and solve for t:

0 = 3t^2 - 6t - 9.

Dividing everything by 3 to simplify, we get:

0 = t^2 - 2t - 3.

Factoring the quadratic equation, we have:

(t - 3)(t + 1) = 0.

The solutions are t = 3 and t = -1.

However, because time cannot be negative in this context, we discard t = -1.

Thus, the particle is at rest at t = 3 seconds.