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 −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.
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Answer 1

To determine the rate of change in the squirrel's height above the ground, we can use the formula for rate of change, which is given by:

\[ \text{Rate of Change} = \frac{\text{Change in Height}}{\text{Change in Time}} \]

1. Calculate the change in height:

   • The initial height of the squirrel is 100 feet (from the perch).
   • The final height of the squirrel (where the acorns are stored) is 45 feet.
   • The change in height = Final height - Initial height = 45 feet - 100 feet = -55 feet.

2. Determine the change in time:

   • The time taken to move from the perch to the acorns is 5.25 seconds.

3. Now calculate the rate of change: \[ \text{Rate of Change} = \frac{-55 \text{ feet}}{5.25 \text{ seconds}} \approx -10.48 \text{ feet/second} \]

The negative sign indicates that the squirrel is descending.

Thus, the quotient that describes the rate of change in the squirrel’s height above the ground is:

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