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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)

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 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.

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.

1 answer

To find the rate of change in the squirrel’s height above the ground, we need to determine how much the squirrel's height changes over the time it takes to reach its acorns.

  1. Calculate the height change:

    • Initial height (perch) = 100 feet
    • Final height (acorns) = 45 feet
    • Change in height = Final height - Initial height
    • Change in height = 45 feet - 100 feet = -55 feet (the height decreases)

  2. Time taken:

    • Time = 5.25 seconds

  3. Calculate the rate of change:

    • Rate of change = Change in height / Time
    • Rate of change = -55 feet / 5.25 seconds ≈ -10.48 feet/second

Thus, the correct interpretation is:

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