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.
-
Calculate \( s(0) \): \[ s(0) = 0^3 - 6(0^2) + 9(0) - 5 = -5 \]
-
Calculate \( s(5) \): \[ s(5) = 5^3 - 6(5^2) + 9(5) - 5 \] \[ = 125 - 150 + 45 - 5 = 15 \]
-
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:
-
For \( t = 1 \): \[ a(1) = 6(1) - 12 = 6 - 12 = -6 , \text{m/s}^2 \]
-
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 \( t = 3 \): \(6 , \text{m/s}^2\)
1 answer