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 of the particle, we start with the equation of motion \( s = t^3 - 27t \). First, let's find the velocity function: 1. Velocity \( v(t) \) is the first derivative of the position \( s \): \[ v(t) = \frac{ds}{dt} = \frac{d}{dt}(t^3 - 27t) = 3t^2 - 27 \] Therefore, \( v(t) = 3t^2 - 27 \). Next, we find the acceleration function: 2. Acceleration \( a(t) \) is the derivative of the velocity \( v(t) \): \[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 27) = 6t \] Therefore, \( a(t) = 6t \, \mathrm{m/s}^2 \). To answer the specific questions: (b) Find the acceleration after 4 seconds: \[ a(4) = 6(4) = 24 \, \mathrm{m/s}^2 \] (c) To find the acceleration when the velocity is 0, we first set \( v(t) = 0 \): \[ 3t^2 - 27 = 0 \implies 3t^2 = 27 \implies t^2 = 9 \implies t = 3 \text{ (since } t \geq 0\text{)} \] Now we use this time to find the acceleration: \[ a(3) = 6(3) = 18 \, \mathrm{m/s}^2 \] So the final answers are: - \( v(t) = 3t^2 - 27 \) - \( a(t) = 6t \, \mathrm{m/s}^2 \) - After 4 seconds: \( 24 \, \mathrm{m/s}^2 \) - When velocity is 0: \( 18 \, \mathrm{m/s}^2 \)