Falls resulting in hip fractures (or head injuries) are a major cause of injury and even death to the
elderly. Typically, the hip’s speed at impact is about 2.0 meters per second. If this can be reduced
to 1.3 meters per second or less, the hip will usually not fracture. One way to do this is by wearing
elastic hip pads.
(a) If a typical pad is 5.0 centimeters thick and compresses by 2.0 centimeters during the impact of
a fall, what acceleration (in meter per second squared and in g’s) does the hip undergo to reduce
its speed to 1.3 meters per second?
(b) The acceleration you found in part (a) may seem like a rather large acceleration, but to fully
assess its effects on the hip, calculate how long it lasts.
To calculate the acceleration (in m/s^2) that the hip undergoes during the impact, we can use the following equation:
v^2 = u^2 + 2as
where:
v = final velocity of the hip after compression (1.3 m/s)
u = initial velocity of the hip before compression (2.0 m/s)
a = acceleration
s = distance the hip pad compresses (2.0 cm, or 0.02 m)
(a) Plugging in these values, we can solve for 'a':
(1.3 m/s)^2 = (2.0 m/s)^2 + 2a(0.02 m)
1.69 m^2/s^2 = 4.0 m^2/s^2 + 0.04a
0.04a = -2.31 m^2/s^2
a = -2.31 m^2/s^2 / 0.04
a ≈ -57.75 m^2/s^2
To convert the acceleration to g's, divide it by the acceleration due to gravity (9.8 m/s²):
a_g = a / 9.8
a_g ≈ -5.89 g (where 'g' is the acceleration due to gravity)
Therefore, the hip undergoes an acceleration of approximately -57.75 m/s^2 or -5.89 g's during the impact.
(b) To calculate how long the acceleration lasts, we can use the equation:
v = u + at
where:
t = time
Rearranging the equation, we get:
t = (v - u) / a
Plugging in the values:
t = (1.3 m/s - 2.0 m/s) / -57.75 m/s^2
t ≈ 0.013 seconds
Therefore, the acceleration of approximately -57.75 m/s^2 (or -5.89 g's) lasts for approximately 0.013 seconds.
To answer part (a) of the question, we need to calculate the acceleration experienced by the hip during impact. We can use the equation of motion:
v^2 = u^2 + 2as
where:
v = final velocity (1.3 m/s)
u = initial velocity (2.0 m/s)
a = acceleration
s = displacement (5.0 cm = 0.05 m)
Rearranging the equation, we get:
a = (v^2 - u^2) / (2s)
Substituting the given values:
a = (1.3^2 - 2^2) / (2 * 0.05)
a = (1.69 - 4) / 0.1
a = -2.31 m/s^2
So, the acceleration experienced by the hip is -2.31 m/s^2 in the direction opposite to its initial velocity.
To convert the acceleration to g's, we divide it by the acceleration due to gravity (9.8 m/s^2):
acceleration in g's = (-2.31 m/s^2) / (9.8 m/s^2)
acceleration in g's ≈ -0.24 g (where 1 g = 9.8 m/s^2)
Therefore, the acceleration experienced by the hip during impact is approximately -0.24 g.
Moving on to part (b) of the question, we need to calculate the duration of the acceleration. We can use the equation of motion:
v = u + at
where:
t = time
v = final velocity (1.3 m/s)
u = initial velocity (2.0 m/s)
a = acceleration
Rearranging the equation, we get:
t = (v - u) / a
Substituting the given values:
t = (1.3 - 2) / -2.31
t ≈ 0.33 seconds
Therefore, the duration of the acceleration is approximately 0.33 seconds.