Question

Consider a particle moving along the \(x\) -axis where \(x(t)\) is the position of the particle at time \(t, x^{\prime}(t)\) is its velocity, and \(x^{\prime \prime}(t)\) is its acceleration. \(x(t)=t^{3}-6 t^{2}+9 t-2, \quad 0 \leq t \leq 5\) (a) Find the velocity and acceleration of the particle. (b) Find the open \(t\) -intervals on which the particle is moving to the right. (c) Find the velocity of the particle when the acceleration is \(0 .\)

Short answer

The velocity function is \(x'(t) = 3t^2 - 12t + 9\). The acceleration function is \(x''(t) = 6t - 12\). The particle moves to the right for \(t > 1\) and \(t < 3\). The velocity of the particle when the acceleration is zero is \(-3\).

Step-by-step solution

Find the velocity

The velocity function is the first derivative of the position function, so differentiate the given position function with respect to \(t\). The derivative of \(t^{3}-6 t^{2}+9 t-2\) is \(3t^2 - 12t + 9\). Therefore, the velocity of the particle, \(x'(t)\), is \(3t^2 - 12t + 9\).

Find the acceleration

The acceleration function is the derivative of the velocity function, so differentiate the function \(3t^2 - 12t + 9\) with respect to \(t\). The derivative is \(6t - 12\). Therefore, the acceleration of the particle, \(x''(t)\), is \(6t - 12\).

Find when the particle is moving to the right

The particle is moving to the right if its velocity is positive. By solving the inequality \(3t^2 - 12t + 9 > 0\), we find that \(t > 1\) and \(t < 3\) are the intervals on which the particle is moving to the right.

Find the velocity when acceleration is zero

The acceleration is zero when \(6t - 12 = 0\). Solving for \(t\), we get \(t=2\). Substituting \(t=2\) into the velocity function \(3t^2 - 12t + 9\) we get \(x'(2)= -3\). So the velocity of the particle when the acceleration is zero is \(-3\).