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find where the object is speeding up and slowing down

The position function of the object is given as:

\[
s(t) = 45t^2 - 109t + 22
\]

To determine where the object is speeding up and slowing down, we need to analyze the velocity and acceleration of the object. Here's the process:

### Step 1: Find the velocity function
The velocity is the first derivative of the position function with respect to time \( t \):

\[
v(t) = \frac{ds(t)}{dt} = \frac{d}{dt}(45t^2 - 109t + 22)
\]

\[
v(t) = 90t - 109
\]

### Step 2: Find the acceleration function
The acceleration is the derivative of the velocity function:

\[
a(t) = \frac{dv(t)}{dt} = \frac{d}{dt}(90t - 109)
\]

\[
a(t) = 90
\]

### Step 3: Analyze speeding up and slowing down
- The object is **speeding up** when the velocity and acceleration have the same sign (either both positive or both negative).
- The object is **slowing down** when the velocity and acceleration have opposite signs.

Since the acceleration \( a(t) = 90 \) is constant and positive, we only need to analyze the velocity function \( v(t) = 90t - 109 \).

1. **Set the velocity equal to zero** to find when the object changes direction:

\[
v(t) = 0 \quad \Rightarrow \quad 90t - 109 = 0
\]
\[
t = \frac{109}{90} \approx 1.211
\]

### Step 4: Interpret the results
- For \( t < 1.211 \), \( v(t) < 0 \) (the object is moving in the negative direction).
- For \( t > 1.211 \), \( v(t) > 0 \) (the object is moving in the positive direction).

Since the acceleration is always positive, the object is:
- **Slowing down** when \( t < 1.211 \) (because \( v(t) \) is negative and \( a(t) \) is positive).
- **Speeding up** when \( t > 1.211 \) (because \( v(t) \) is positive and \( a(t) \) is positive).

### Conclusion:
- The object is **slowing down** in the interval \( [0, 1.211] \).
- The object is **speeding up** in the interval \( [1.211, 5] \).

Would you like more details on any part of this process?

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Here are five related questions for further exploration:
1. How would the analysis change if the acceleration was not constant?
2. What is the object's velocity at \( t = 0 \) and \( t = 5 \)?
3. How far has the object traveled by \( t = 5 \)?
4. What would happen if the position function had a cubic term?
5. How would you find the maximum or minimum position of the object?

**Tip:** To better understand motion, always analyze both velocity and acceleration functions. Velocity tells you the direction and speed of motion, while acceleration reveals changes in that speed.

Find where the object is speeding up and slowing down

Find out where an object is speeding up or slowing down using a position function. This problem covers key calculus concepts, including velocity and acceleration, and provides a step-by-step solution for determining intervals of speeding up and slowing down based on the velocity and acceleration signs. Suitable for high school and university students.

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