Answer:
While falling, the magnitude of the acceleration of the egg is 9.82 m/s²
While stopping, the magnitude of the deceleration of the egg is 79.3 m/s²
Explanation:
Hi there!
The equation of velocity of the falling egg is the following:
v = v0 + a · t
Where:
v = velocity at time t.
v0 = initial velocity.
a = acceleration.
t = time
Let´s calculate the acceleration of the egg while falling. Notice that the result should be the acceleration of gravity, ≅ 9.8 m/s².
v = v0 + a · t
11.1 m/s = 0 m/s + a · 1.13 s (since the egg is dropped, the initial velocity is zero). Solving for "a":
11.1 m/s / 1.13 s = a
a = 9.82 m/s²
While falling, the magnitude of the acceleration of the egg is 9.82 m/s²
Now, using the same equation, let´s find the acceleration of the egg while stopping. We know that at t = 0.140 s after touching the ground, the velocity of the egg is zero. We also know that the velocity of the egg before hiiting the ground is 11.1 m/s, then, v0 = 11.1 m/s:
v = v0 + a · t
0 = 11.1 m/s + a · 0.140 s
-11.1 m/s / 0.140 s = a
a = -79.3 m/s²
While stopping, the magnitude of the deceleration of the egg is 79.3 m/s²
