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 squirrels height above the ground. Give your answer to the nearest hundredth
To find the rate of change in the squirrel's height above the ground, we can divide the change in height by the change in time.
The initial height of the squirrel above the ground is 100 feet, and it moves down to a height of 45 feet above the ground. Thus, the change in height is 100 - 45 = 55 feet.
The change in time is given as 5.25 seconds.
Therefore, the rate of change in the squirrel's height above the ground is 55 feet / 5.25 seconds = 10.48 feet per second.
Rounded to the nearest hundredth, the rate of change in the squirrel's height above the ground is 10.48 feet/second.
To interpret the quotient in this scenario, we need to calculate the rate of change in the squirrel's height above the ground.
Rate of change can be found by dividing the change in height by the time taken.
Change in height = 100 ft (initial height) - 45 ft (final height) = 55 ft
Time taken = 5.25 seconds
Rate of change = Change in height / Time taken
Rate of change = 55 ft / 5.25 seconds ≈ 10.48 ft/s
Therefore, the rate of change in the squirrel's height above the ground is approximately 10.48 ft/s.
To find the rate of change in the squirrel's height above the ground, we need to determine the distance it has traveled vertically and divide it by the time it took.
Given that the squirrel starts on a perch 100 feet above the ground and moves down to its stored acorns in 5.25 seconds, the distance it has traveled vertically is 100 - 45 = 55 feet (since it moves down).
To calculate the rate of change, we divide the distance traveled (55 feet) by the time it took (5.25 seconds):
Rate of change = distance traveled / time taken
= 55 feet / 5.25 seconds
Calculating this division gives us approximately 10.48 feet per second.
Therefore, the rate of change in the squirrel's height above the ground is approximately 10.48 feet per second.