Question

The position of a particle at time t sec is s = t^3-6t^2+9t-5 Find the displacement during the first 5 seconds, find the average velocity during the first 5 seconds and find the acceleration for each time the particles velocity is zero

Answer 1

To analyze the motion of the particle defined by the position function \( s(t) = t^3 - 6t^2 + 9t - 5 \), we'll go through the steps you've provided.

### Step 1: Find the displacement during the first 5 seconds.

The displacement during a time interval is given by the difference between the position at the end of the interval and the position at the beginning of the interval.

1. Calculate \( s(0) \):
\[
s(0) = 0^3 - 6(0^2) + 9(0) - 5 = -5
\]

2. Calculate \( s(5) \):
\[
s(5) = 5^3 - 6(5^2) + 9(5) - 5
\]
\[
= 125 - 150 + 45 - 5 = 15
\]

3. Calculate displacement:
\[
\text{Displacement} = s(5) - s(0) = 15 - (-5) = 15 + 5 = 20
\]

### Step 2: Find the average velocity during the first 5 seconds.

The average velocity over a time interval is given by the total displacement divided by the length of the time interval.

\[
\text{Average Velocity} = \frac{\text{Displacement}}{\Delta t} = \frac{20}{5} = 4 \, \text{m/s}
\]

### Step 3: Find the acceleration for each time the particle's velocity is zero.

First, we need to find the velocity function, which is given by the derivative of the position function:
\[
v(t) = \frac{ds}{dt} = \frac{d}{dt}(t^3 - 6t^2 + 9t - 5) = 3t^2 - 12t + 9
\]

To find when the velocity is zero, set \( v(t) = 0 \):
\[
3t^2 - 12t + 9 = 0
\]
Dividing the equation by 3:
\[
t^2 - 4t + 3 = 0
\]

Factoring:
\[
(t - 1)(t - 3) = 0
\]
Thus,
\[
t = 1 \quad \text{and} \quad t = 3
\]

Next, we calculate the acceleration, which is the derivative of the velocity function, \( a(t) = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 12t + 9) \):
\[
a(t) = 6t - 12
\]

Now, we can find the acceleration at the points where the velocity is zero:

1. For \( t = 1 \):
\[
a(1) = 6(1) - 12 = 6 - 12 = -6 \, \text{m/s}^2
\]

2. For \( t = 3 \):
\[
a(3) = 6(3) - 12 = 18 - 12 = 6 \, \text{m/s}^2
\]

### Summary of Results:
- **Displacement during the first 5 seconds**: 20 meters
- **Average velocity during the first 5 seconds**: 4 m/s
- **Acceleration when velocity is zero**:
- At \( t = 1 \): \(-6 \, \text{m/s}^2\)
- At \( t = 3 \): \(6 \, \text{m/s}^2\)