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 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.
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 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.

Answer 1

To calculate the rate of change in the squirrel’s height above the ground, we start by determining the change in height and then divide by the time taken.

1. Initial height: 100 feet (the perch)
2. Final height: 45 feet (the hole)

The change in height (Δh) can be calculated as: \[ \Delta h = \text{Final height} - \text{Initial height} = 45 , \text{feet} - 100 , \text{feet} = -55 , \text{feet} \]

Next, we divide the change in height by the time taken (5.25 seconds): \[ \text{Rate of change} = \frac{\Delta h}{\Delta t} = \frac{-55 , \text{feet}}{5.25 , \text{seconds}} \approx -10.48 , \text{feet/second} \]

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

The correct interpretation from the responses is:

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.