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find where the object is speeding up and slowing down. Solve the problem following the method in the written example.

Here is the solution to your problem:

1. **Velocity function (first derivative)**: \( v(t) = 90t - 109 \)
2. **Acceleration function (second derivative)**: \( a(t) = 90 \) (constant)

3. **Turning point**: The object changes its behavior at \( t = \frac{109}{90} \approx 1.21 \).

**Intervals for speeding up and slowing down:**
- The object is **slowing down** on the interval \( (0, \frac{109}{90}) \) or approximately \( (0, 1.21) \).
- The object is **speeding up** on the interval \( (\frac{109}{90}, 5) \) or approximately \( (1.21, 5) \).

Would you like further details on how this solution was derived?

Here are five related questions:
1. How do we determine when an object is speeding up or slowing down using calculus?
2. Why is the acceleration constant in this case?
3. What are the conditions for an object to be speeding up or slowing down based on velocity and acceleration?
4. How do turning points affect the motion of an object?
5. How would the intervals change if the acceleration were not constant?

**Tip**: Remember that when velocity and acceleration have the same sign, the object speeds up; when they have opposite signs, the object slows 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.

Learn how to find when an object is speeding up or slowing down using the position function 45t^2 - 109t + 22. Follow a detailed step-by-step solution that involves calculating the velocity and acceleration and determining intervals of motion. Suitable for high school students studying calculus.

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