Question

An egg drops from a second-story window, taking 1.13 s to fall and reaching a speed of 11.1 m/s just before hitting the ground. On contact with the ground, the egg stops completely in 0.140 s. Calculate the average magnitudes of its acceleration while falling and while stopping.
m/s2 (acceleration while falling)
m/s2 (deceleration while stopping)

Answer 1

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