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We need to convert the acceleration from **yards/minute²** to **feet/second²**.

Given:
- \( 1 \text{ yard} = 3 \text{ feet} \)
- \( 1 \text{ minute} = 60 \text{ seconds} \)

The acceleration given is \( 9 \, \text{yards}/\text{minute}^2 \).

### Step 1: Convert yards to feet
To convert the value from yards to feet:
\[
9 \, \text{yards}/\text{minute}^2 \times 3 \, \text{feet}/\text{yard} = 27 \, \text{feet}/\text{minute}^2
\]

### Step 2: Convert minutes² to seconds²
Now, convert the time unit from **minutes²** to **seconds²**. Since \( 1 \, \text{minute} = 60 \, \text{seconds} \), we square this value:
\[
1 \, \text{minute}^2 = (60 \, \text{seconds})^2 = 3600 \, \text{seconds}^2
\]
Now convert:
\[
27 \, \text{feet}/\text{minute}^2 \times \frac{1}{3600} \, \text{minute}^2/\text{second}^2 = \frac{27}{3600} \, \text{feet}/\text{second}^2
\]

### Final Answer:
\[
\frac{27}{3600} \, \text{feet}/\text{second}^2
\]
This is the answer in feet per second squared without simplifying.

Would you like more details or clarifications?

Here are 5 related questions:
1. How would you simplify the result further if required?
2. What would happen if the units were given in miles instead of yards?
3. How does the conversion change if acceleration is given in meters per second squared?
4. What is the physical interpretation of acceleration in these units?
5. Why is squaring the time unit important during conversion?

**Tip:** Always ensure both distance and time units are consistently converted before calculating.

Write the acceleration 9 yards/minute² in units that involve only feet and seconds. Do not simplify your answer.

Learn how to convert acceleration from 9 yards/minute² to feet/second² by using basic unit conversion principles. This problem covers acceleration unit conversion from yards to feet and minutes to seconds. Suitable for students in Grades 8-10.

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