Question

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. (a) If a typical pad is 5.0 cm thick and compresses by 2.0 cm during the impact of a fall, what constant acceleration (in m/s2 and in g’s) does the hip undergo to reduce its speed from 2.0 m/s to 1.3 m/s? (b) The acceleration you found in part (a) may seem rather large, but to assess its effects on the hip, calculate how long it lasts.

Answer 1

Answer:

-5.8868501529 m/s² or -5.8868501529g

0.118909090909 s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{1.3^2-2^2}{2\times 0.02}\\\Rightarrow a=-5.8868501529\ m/s^2$

Dividing by g

$\dfrac{a}{g}=\dfrac{-57.75}{9.81}\\\Rightarrow a=-5.8868501529g$

The acceleration is -5.8868501529 m/s² or -5.8868501529g

$v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{1.3-2}{-5.8868501529}\\\Rightarrow t=0.118909090909\ s$

The time taken is 0.118909090909 s