The function \( s(t) \) describes the motion of a particle along a line. \[ s(t)=4 t-t^{2} \] (a) Find the velocity function \( v(t) \) of the particle at any time \( t \geq 0 \). \[ v(t)=\text { (b) Identify the time interval(s) on which the particle is moving in a positive direction. (Enter your answer using interval notation.) } \] (c) Identify the time interval(s) on which the particle is moving in a negative direction. (Enter your answer using interval notation.) (d) Identify the time(s) at which tha particle changes direction. (Enter your answers as a comma-separated list.) \( t=\square \)

To find the velocity function \( v(t) \), we need to differentiate the position function \( s(t) \): \[ v(t) = \frac{ds(t)}{dt} = \frac{d}{dt}(4t - t^2) = 4 - 2t. \] (a) Thus, the velocity function is: \[ v(t) = 4 - 2t. \] Now, let's identify the intervals of movement: (b) The particle is moving in a positive direction when \( v(t) > 0 \): \[ 4 - 2t > 0 \] \[ 2t < 4 \] \[ t < 2. \] In interval notation, the particle moves positively on: \[ (0, 2). \] (c) The particle is moving in a negative direction when \( v(t) < 0 \): \[ 4 - 2t < 0 \] \[ 2t > 4 \] \[ t > 2. \] In interval notation, the particle moves negatively on: \[ (2, \infty). \] (d) The particle changes direction when \( v(t) = 0 \): Setting the velocity function equal to zero: \[ 4 - 2t = 0 \] \[ 2t = 4 \] \[ t = 2. \] So, the particle changes direction at: \( t = 2 \).