Question
A squirrel has stored its corn in a hole that is 45 feet from the ground in a tall tree. The school starts on a purge 100 feet above the ground. The squirrel moves from the perch down to its start acorn in 5.2 5 seconds and interpret the quote to describe the rate of the change in the squirrel height above the ground your answer to the nearest hundred.
Answers
Answered by
GPT-4o mini
To find the rate of change in the squirrel's height above the ground, we can use the formula for the rate of change, which is the change in height divided by the change in time.
1. **Initial height of the squirrel at the perch**: 100 feet
2. **Final height of the squirrel at the corn**: 45 feet
3. **Change in height**: 100 feet - 45 feet = 55 feet
4. **Time taken**: 5.25 seconds (note that "5.2 5 seconds" seems to include an error; we'll interpret this as 5.25 seconds for clarity.)
Now, we can calculate the rate of change:
\[
\text{Rate of change} = \frac{\text{Change in height}}{\text{Time}} = \frac{55 \text{ feet}}{5.25 \text{ seconds}} \approx 10.48 \text{ feet/second}
\]
Rounding this to the nearest hundredth gives:
\[
\text{Rate of change} \approx 10.48 \text{ feet/second}
\]
Thus, the rate of change in the squirrel's height above the ground as it descends to retrieve its corn is approximately **10.48 feet per second**.
1. **Initial height of the squirrel at the perch**: 100 feet
2. **Final height of the squirrel at the corn**: 45 feet
3. **Change in height**: 100 feet - 45 feet = 55 feet
4. **Time taken**: 5.25 seconds (note that "5.2 5 seconds" seems to include an error; we'll interpret this as 5.25 seconds for clarity.)
Now, we can calculate the rate of change:
\[
\text{Rate of change} = \frac{\text{Change in height}}{\text{Time}} = \frac{55 \text{ feet}}{5.25 \text{ seconds}} \approx 10.48 \text{ feet/second}
\]
Rounding this to the nearest hundredth gives:
\[
\text{Rate of change} \approx 10.48 \text{ feet/second}
\]
Thus, the rate of change in the squirrel's height above the ground as it descends to retrieve its corn is approximately **10.48 feet per second**.