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.
2^2 - 1.3^2 = 2a(.03) (meters). Compare to 9.8.
Use any other equation of motion to find t.
To answer part (a) of the question, we need to determine the acceleration experienced by the hip during the impact. We can use the equations of motion to calculate this.
Let's start by converting the thickness and compression of the pad to meters.
Thickness of the pad = 5.0 centimeters = 0.05 meters
Compression of the pad = 2.0 centimeters = 0.02 meters
Since the hip's speed is reduced from 2.0 m/s to 1.3 m/s, the change in velocity is:
Δv = 1.3 m/s - 2.0 m/s = -0.7 m/s (negative because the velocity decreases)
Now, we can use the equation of motion:
Δv = a * Δt
Where:
a is the acceleration
Δt is the time interval
Rearranging the equation, we have:
a = Δv / Δt
We can find the time interval Δt by considering the compression of the pad and assuming a uniform deceleration.
The distance (d) traveled during deceleration can be given by:
d = 0.02 m
We can find Δt using the equation:
d = (1/2) * a * Δt^2
Rearranging the equation, we get:
Δt = √(2 * d / a)
Now we can plug in the values:
Δt = √(2 * 0.02 m / a)
Solving for a:
a = 0.02 m / (√(2 * 0.02 m / a))
Simplifying further:
a^2 = 2 * 0.02 m
a = √(2 * 0.02 m)
a ≈ 0.2 m/s^2
To find the acceleration in g's, we can divide the acceleration by the acceleration due to gravity (g = 9.8 m/s^2):
a_g = a / g
a_g = 0.2 m/s^2 / 9.8 m/s^2
a_g ≈ 0.020 g
Therefore, the acceleration experienced by the hip during the impact is approximately 0.2 m/s^2 or 0.020 g.
To answer part (b) of the question, we need to calculate the time interval (Δt) during which the acceleration occurs. We have already found this time interval to calculate the acceleration.
Using the previously calculated value of Δt:
Δt = √(2 * 0.02 m / 0.2 m/s^2)
Δt = √(0.04 / 0.2)
Δt = √0.2
Δt ≈ 0.447 seconds
Therefore, the acceleration lasts for approximately 0.447 seconds.