19. [-14 Paints] DETAILS MY NOTES SCALCET9 3.1.053. The equation of motion of a particle is \( s=t^{3}-27 t \), where \( s \) is measured in meters and \( t \) is in seconds. (Assume \( t \geq 0 \). ) (a) Find the velocity and acceleration as functions of \( t \). \( v(t)=\square \) \( a(t)=\square \mathrm{m} / \mathrm{s}^{2} \) (b) Find the acceleration, in \( \mathrm{m} / \mathrm{s}^{2} \), after 4 seconds. (c) Find the acceleration, in \( \mathrm{m} / \mathrm{s}^{2} \), when the velocity is 0 . (c) \( \mathrm{m} / \mathrm{s}^{2} \)

To find the velocity and acceleration functions of the particle, we start with the position function \( s(t) = t^3 - 27t \). (a) The velocity \( v(t) \) is the first derivative of the position function with respect to time \( t \): \[ v(t) = \frac{ds}{dt} = 3t^2 - 27. \] The acceleration \( a(t) \) is the derivative of the velocity function: \[ a(t) = \frac{dv}{dt} = 6t. \] (b) To find the acceleration after 4 seconds: \[ a(4) = 6 \times 4 = 24 \, \mathrm{m/s^2}. \] (c) To find when the velocity is 0: Setting \( v(t) = 0 \): \[ 3t^2 - 27 = 0 \Rightarrow t^2 = 9 \Rightarrow t = 3. \] Now plug \( t = 3 \) into the acceleration function: \[ a(3) = 6 \times 3 = 18 \, \mathrm{m/s^2}. \] Thus, the answers are: \( v(t)= 3t^2 - 27 \) \( a(t)= 6t \, \mathrm{m/s^2} \) For the specific values: Acceleration after 4 seconds: \( 24 \, \mathrm{m/s^2} \) Acceleration when velocity is 0: \( 18 \, \mathrm{m/s^2} \)