A squirrel has stored its acorns in a hole that is 45 feet from the ground in a tall tree. The squirrel starts on a perch 100 feet above the ground. The squirrel moves from the perch down to its stored acorns in 5.25 seconds. Interpret the quotient to describe the rate of change in the squirrel’s height above the ground. Give your answer to the nearest hundredth.(1 point)

Question

Responses

The quotient that describes the rate of change in the squirrel’s height above the ground is 19.05 feet/second.
The quotient that describes the rate of change in the squirrel’s height above the ground is 19.05 feet/second.

The quotient that describes the rate of change in the squirrel’s height above the ground is 10.48 feet/second.
The quotient that describes the rate of change in the squirrel’s height above the ground is 10.48 feet/second.

The quotient that describes the rate of change in the squirrel’s height above the ground is −19.05
feet/second.
The quotient that describes the rate of change in the squirrel’s height above the ground is negative 19.05 feet/second.

The quotient that describes the rate of change in the squirrel’s height above the ground is −10.48
feet/second.

Answer 1

To find the rate of change in the squirrel’s height above the ground, we first need to determine the total height the squirrel is descending and the time it takes.

1. Initial height: The squirrel starts at 100 feet.
2. Final height: The squirrel's acorns are stored at 45 feet.
3. Height change = Initial height - Final height = 100 feet - 45 feet = 55 feet.
4. Time taken: 5.25 seconds.

Now we can calculate the rate of change using the formula:

\[ \text{Rate of change} = \frac{\text{Change in height}}{\text{Time}} = \frac{-55 \text{ feet}}{5.25 \text{ seconds}} \]

Calculating this gives:

\[ \text{Rate of change} = -10.48 \text{ feet/second} \]

So, the quotient that describes the rate of change in the squirrel’s height above the ground is −10.48 feet/second.

Thus, the correct response is:

The quotient that describes the rate of change in the squirrel’s height above the ground is −10.48 feet/second.