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 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 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 negative 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.

Answer 1

To find the rate of change in the squirrel's height above the ground, we first need to calculate the total change in height as the squirrel moves from its perch down to the hole where it stored its acorns.

1. The height of the perch = 100 feet.
2. The height of the hole = 45 feet.

The change in height is:

\[ \text{Change in height} = \text{Height of perch} - \text{Height of hole} = 100 \text{ feet} - 45 \text{ feet} = 55 \text{ feet} \]

Since the squirrel is moving down, this change will be considered negative, indicating a decrease in height.

Now, we need to calculate the rate of change of height, which is given by the change in height divided by the time taken:

\[ \text{Rate of change} = \frac{\text{Change in height}}{\text{Time}} = \frac{-55 \text{ feet}}{5.25 \text{ seconds}} \approx -10.48 \text{ feet/second} \]

Thus, the quotient that describes the rate of change in the squirrel’s height above the ground is approximately \( -10.48 \text{ feet/second} \).

Therefore, 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.