Question
A score last stored acorns in it all that’s 45 feet above the ground and I told the squirrel starts on a purchase 100 feet above the ground the squirrel from the perch down to its stored a coordinate 5.25 seconds interpret the question and describe the rate of that change the score side above the ground answer the nearest hundred
Answers
Answered by
GPT-4o mini
To interpret the question, it appears that you have a scenario involving a squirrel that starts at a height of 100 feet above the ground and moves down to a stored location of acorns that is 45 feet above the ground. The time it takes for the squirrel to make this descent is 5.25 seconds.
To describe the rate of change in this case, we can calculate the vertical displacement of the squirrel and divide it by the time it took to cover that distance.
1. **Initial height (starting position)**: 100 feet
2. **Final height (location of stored acorns)**: 45 feet
3. **Vertical displacement**:
\[
\text{Vertical Displacement} = \text{Initial height} - \text{Final height} = 100 \text{ feet} - 45 \text{ feet} = 55 \text{ feet}
\]
4. **Time taken**: 5.25 seconds
5. **Rate of change (velocity)**:
\[
\text{Rate of change} = \frac{\text{Vertical Displacement}}{\text{Time}} = \frac{55 \text{ feet}}{5.25 \text{ seconds}} \approx 10.48 \text{ feet/second}
\]
Rounding this to the nearest hundredth gives us a rate of change of approximately **10.48 feet per second** downward. This represents how quickly the squirrel is descending toward the stored acorns.
To describe the rate of change in this case, we can calculate the vertical displacement of the squirrel and divide it by the time it took to cover that distance.
1. **Initial height (starting position)**: 100 feet
2. **Final height (location of stored acorns)**: 45 feet
3. **Vertical displacement**:
\[
\text{Vertical Displacement} = \text{Initial height} - \text{Final height} = 100 \text{ feet} - 45 \text{ feet} = 55 \text{ feet}
\]
4. **Time taken**: 5.25 seconds
5. **Rate of change (velocity)**:
\[
\text{Rate of change} = \frac{\text{Vertical Displacement}}{\text{Time}} = \frac{55 \text{ feet}}{5.25 \text{ seconds}} \approx 10.48 \text{ feet/second}
\]
Rounding this to the nearest hundredth gives us a rate of change of approximately **10.48 feet per second** downward. This represents how quickly the squirrel is descending toward the stored acorns.