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A coyote chase the rabbit 65.0 m West to a bush and then chase the rabbit another 23.2 m West before catching it if the entire Chase lasted 97 seconds what was the average velocity of the chase.

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Final answer:

To calculate the average velocity of the coyote's chase, add the total displacements (65.0 m + 23.2 m = 88.2 m West) and divide by the total time (97 s) to get an average velocity of approximately 0.909 m/s West.

Step-by-step explanation:

The question addresses the concept of average velocity in physics. To find the average velocity of the coyote's chase, we need to calculate the total displacement of the coyote and divide it by the total time taken for the chase.

First, we calculate the total displacement. Since the coyote chases the rabbit first 65.0 m West and then another 23.2 m West, the total displacement (d) is:

d = 65.0 m + 23.2 m

d = 88.2 m West

Next, we use the formula for average velocity (vavg), which is:

vavg = total displacement / total time

The total displacement is 88.2 m West, and the chase lasted 97 seconds.

Therefore, the average velocity (vavg) is:

vavg = 88.2 m / 97 s

vavg ≈8 0.909 m/s West

This represents the average velocity of the coyote during the chase.

User Alvin Theodora
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