To find the time when the velocity and acceleration of the particle are zero based on the given position function \( s(t) = t^3 - 3t^2 + 9t + 1 \), we first need to calculate the expressions for velocity and acceleration.
Step 1: Find the velocity \( v(t) \)
The velocity is the derivative of the position function with respect to time \( t \):
\[ v(t) = \frac{ds}{dt} = \frac{d}{dt}(t^3 - 3t^2 + 9t + 1) \]
Calculating the derivative:
\[ v(t) = 3t^2 - 6t + 9 \]
Step 2: Find the time when velocity is zero
To find when the velocity is zero, set \( v(t) = 0 \):
\[ 3t^2 - 6t + 9 = 0 \]
Dividing the entire equation by 3:
\[ t^2 - 2t + 3 = 0 \]
Now, we can use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -2, c = 3 \):
\[ t = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} \] \[ = \frac{2 \pm \sqrt{4 - 12}}{2} \] \[ = \frac{2 \pm \sqrt{-8}}{2} \] \[ = \frac{2 \pm 2i\sqrt{2}}{2} \] \[ = 1 \pm i\sqrt{2} \]
Since the solutions are complex, the particle's velocity does not equal zero at any real time. Therefore:
\[ \text{Velocity is never zero for real values of } t. \]
Step 3: Find the acceleration \( a(t) \)
The acceleration is the derivative of the velocity function:
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 6t + 9) \]
Calculating the derivative:
\[ a(t) = 6t - 6 \]
Step 4: Find the time when acceleration is zero
Set \( a(t) = 0 \):
\[ 6t - 6 = 0 \]
Solving for \( t \):
\[ 6t = 6 \] \[ t = 1 \]
Summary of Results
- The time when the velocity is zero: Never for real values of \( t \).
- The time when the acceleration is zero: \( t = 1 \).