The function \( s(t) \) describes the motion of a particle along a line. \[ s(t)=t^{3}-8 t^{2}+7 t \] (a) Find the velocity function of the particle at any time \( t \geq 0 \). \[ v(t)=\square \] (b) Identify the time intervals on which the particle is moving in a positive direction. (Enter your answer using interval notation.) \( \square \) (c) Identify the time intervals on which the particle is moving in a negative direction. (Enter your answer using interval notation.) \( \square \) (d) Identify the time(s) at which the particle changes direction. (Enter your answers as a comma-separatad list.) \[ t=\square \]

To solve this problem, we need to analyze the motion of the particle based on the given position function \( s(t) = t^3 - 8t^2 + 7t \). (a) To find the velocity function \( v(t) \), we take the derivative of the position function with respect to time \( t \): \[ v(t) = s'(t) = \frac{d}{dt}(t^3 - 8t^2 + 7t) = 3t^2 - 16t + 7. \] (b) The particle is moving in a positive direction when \( v(t) > 0 \). To find the intervals, we need to determine where \( 3t^2 - 16t + 7 > 0 \). We can start by finding the roots of the equation \( 3t^2 - 16t + 7 = 0 \) using the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 3 \cdot 7}}{2 \cdot 3} = \frac{16 \pm \sqrt{256 - 84}}{6} = \frac{16 \pm \sqrt{172}}{6} = \frac{16 \pm 2\sqrt{43}}{6} = \frac{8 \pm \sqrt{43}}{3}. \] These roots are approximately \( t \approx 0.897 \) and \( t \approx 5.103 \). To find the intervals where \( v(t) > 0 \), we test the intervals around these roots: - For \( t < \frac{8 - \sqrt{43}}{3} \) (approximately 0.897), test \( t = 0 \): \( v(0) = 7 > 0 \) (positive). - For \( t \in \left( \frac{8 - \sqrt{43}}{3}, \frac{8 + \sqrt{43}}{3} \right) \), test \( t = 3 \): \( v(3) = 3(9) - 16(3) + 7 = 27 - 48 + 7 = -14 < 0 \) (negative). - For \( t > \frac{8 + \sqrt{43}}{3} \) (approximately 5.103), test \( t = 6 \): \( v(6) = 3(36) - 16(6) + 7 = 108 - 96 + 7 = 19 > 0 \) (positive). Thus, the particle moves in a positive direction on the intervals \( (0, \frac{8 - \sqrt{43}}{3}) \) and \( (\frac{8 + \sqrt{43}}{3}, \infty) \). (c) The particle is moving in a negative direction when \( v(t) < 0 \). Based on our intervals from part (b), the particle moves in a negative direction on the interval \( \left(\frac{8 - \sqrt{43}}{3}, \frac{8 + \sqrt{43}}{3}\right) \). (d) The particle changes direction when \( v(t) = 0 \), which occurs at the roots we calculated earlier: \[ t = \frac{8 - \sqrt{43}}{3}, \frac{8 + \sqrt{43}}{3}. \] Now we can summarize our answers: (a) \[ v(t) = 3t^2 - 16t + 7. \] (b) \( (0, \frac{8 - \sqrt{43}}{3}) \cup (\frac{8 + \sqrt{43}}{3}, \infty) \). (c) \( \left( \frac{8 - \sqrt{43}}{3}, \frac{8 + \sqrt{43}}{3} \right) \). (d) \[ t = \frac{8 - \sqrt{43}}{3}, \frac{8 + \sqrt{43}}{3}. \]