Question

After. Seconds, a particle has covered a distance of s meters, where s= t3-3t3+9t+1 determine de time when 1) velocity is zero 2) Acceleration is zero

Answer 1

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

1. The time when the velocity is zero: **Never for real values of \( t \)**.
2. The time when the acceleration is zero: **\( t = 1 \)**.