1. 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)
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 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 -19.05 feet/second.
The correct quotient that describes the rate of change in the squirrel's height above the ground is -10.48 feet/second.
To find the rate of change in the squirrel's height above the ground, we need to calculate the average vertical velocity. The formula for average velocity is given by:
average velocity = change in distance / change in time
In this case, the squirrel starts at a height of 100 feet and ends at a height of 45 feet, while taking 5.25 seconds to do so.
Change in distance = final distance - initial distance = 45 - 100 = -55 feet (negative because the squirrel is moving downwards)
Change in time = 5.25 seconds
So, the average velocity is:
average velocity = -55 feet / 5.25 seconds = -10.48 feet/second
Therefore, the correct answer is:
The quotient that describes the rate of change in the squirrel’s height above the ground is -10.48 feet/second.