A cat drops from a shelf 3.9 above the floor and lands on all four feet. His legs bring him to a stop in a distance of 12 .Calculate his acceleration (assumed constant) while he is stopping, in .

2.2

To calculate the cat's acceleration while stopping, we can use the kinematic equation:

v^2 = u^2 + 2as

where:
v = final velocity (0 m/s, as the cat comes to a stop)
u = initial velocity (unknown)
a = acceleration (unknown)
s = displacement (12 m)

Since the cat falls from rest, its initial velocity, u, would be 0 m/s. However, we need to consider that the cat is dropping from a height of 3.9 m. We can calculate the initial velocity just before the cat lands using the equation of motion for free-falling objects:

v^2 = u^2 + 2as

Here, v = unknown, u = 0 m/s, a = 9.8 m/s^2 (acceleration due to gravity), and s = 3.9 m. Plugging in these values, we can solve for v:

v^2 = 0^2 + 2 * 9.8 * 3.9
v^2 = 0 + 76.44
v^2 = 76.44
v ≈ √76.44
v ≈ 8.74 m/s

Now that we know the initial velocity just before the cat lands is approximately 8.74 m/s, we can substitute this value along with the other known values (u, v, and s) into the first kinematic equation to calculate the acceleration:

0^2 = (8.74)^2 + 2 * a * 12

0 = 76.44 + 24a

Rearranging the equation:

24a = -76.44

a = -76.44 / 24

a ≈ -3.18 m/s^2

Therefore, the cat's acceleration while stopping is approximately -3.18 m/s^2. Note that the negative sign indicates that the acceleration is in the opposite direction of the cat's initial motion.