Reducing Hip Fracture Risk with Elastic Hip Pads: What Acceleration is Needed?

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Falls leading to hip fractures in the elderly can be mitigated by using elastic hip pads, which can reduce the impact speed from 2.0 m/s to 1.3 m/s or less. The discussion centers on calculating the necessary acceleration to achieve this reduction, given a 5.0 cm thick pad that compresses by 2.0 cm during impact. The relevant equation for this calculation is v^2 = v0^2 + 2a(x-x0). Clarifications were made regarding the correct values for compression in the equation, emphasizing the importance of accurate measurements. Overall, the conversation highlights the potential effectiveness of elastic hip pads in preventing serious injuries from falls.
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Homework Statement



Falls resulting in hip fractures are a major cause of injury and even death to the elderly. Typically, the hip’s speed at impact is about 2.0 m/s. If this can be reduced to 1.3 m/s or less, the hip will usually not fracture. One way to do this is by wearing elastic hip pads.

If a typical pad is 5.0 cm thick and compresses by 2.0 cm during the impact of a fall, what acceleration (in m/s^2) does the hip undergo to reduce its speed to 1.3 m/s?

Homework Equations



v^2 = v0^2 + 2a(x-x0)

The Attempt at a Solution



My attempt is attached below. Thanks everybody for the help :)
 

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Does it compress to 2cm or by 2cm?
 
Oh it says by. So you're sayin my (x-x0) should be .02 not .03.
 
Correct :)
 
Yep that was the problem. Thanks man you've been helping me all day.
 
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