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We are given the position function \( s(t) = 45t^2 - 109t + 22 \) for an object moving along a line, and we need to determine where the object is speeding up and slowing down within the domain \( [0, 5] \).

### Step 1: Velocity and Acceleration Functions
To find when the object is speeding up or slowing down, we need to consider the velocity and acceleration functions. The object is speeding up when the velocity and acceleration have the same sign, and slowing down when they have opposite signs.

1. **Velocity \( v(t) \)** is the first derivative of the position function:
   \[
   v(t) = \frac{d}{dt} \left( 45t^2 - 109t + 22 \right) = 90t - 109
   \]
   
2. **Acceleration \( a(t) \)** is the second derivative of the position function, or the derivative of the velocity function:
   \[
   a(t) = \frac{d}{dt} (90t - 109) = 90
   \]
   Since \( a(t) = 90 \), acceleration is constant and positive.

### Step 2: Solve for Critical Points
The object changes direction (i.e., velocity is zero) when \( v(t) = 0 \). To find the critical points:
\[
90t - 109 = 0 \quad \Rightarrow \quad t = \frac{109}{90} \approx 1.211
\]
This is the time when the velocity changes sign.

### Step 3: Analyze Intervals
Now, we analyze the sign of the velocity \( v(t) \) over the interval \( [0, 5] \):
- For \( t < 1.211 \), \( v(t) = 90t - 109 \) is negative (since \( 90(0) - 109 = -109 \)).
- For \( t > 1.211 \), \( v(t) = 90t - 109 \) is positive (since \( 90(5) - 109 = 341 \)).

Since acceleration is positive (and constant), the object is:
- **Slowing down** when the velocity is negative, i.e., on \( [0, 1.211] \).
- **Speeding up** when the velocity is positive, i.e., on \( [1.211, 5] \).

### Final Answer
- The object is **speeding up** on \( [1.211, 5] \).
- The object is **slowing down** on \( [0, 1.211] \).

Would you like more details on this explanation, or have any other questions?

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Here are 5 related questions to explore further:
1. How does constant acceleration affect motion over time?
2. What would happen if the acceleration were negative instead of positive?
3. Can the object ever stop if the acceleration is constant and positive?
4. How would you compute the displacement over a given time interval?
5. What are the physical meanings of velocity and acceleration in real-world scenarios?

**Tip:** When solving motion problems, always consider the relationship between velocity and acceleration to determine speeding up or slowing down.

An object moves in 1 dimension along a line according to the position function 45t^2 − 109t + 22 with domain [0,5]. Find where the object is speeding up and slowing down. Enter your answers in interval notation with exact, simplified numbers.

Explore how to determine when an object is speeding up or slowing down using the position function 45t^2 - 109t + 22. This problem includes finding the velocity, acceleration, and analyzing intervals for speeding up and slowing down. Suitable for college or advanced high school calculus students.

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