Please help me calculate the the acceleration in a fall?

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. I this can be reduced to 1.3 m/s or less, the hip will not fracture. One way to do the 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 acceleration in m/s^2 and g's) does the hip undergo to reduce its speed to 1.3 m/s? (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.

Please help me. I have been trying to work on this and just can't figure it out.

Answer 1

One bit of data you are missing is the distance of the fall. I will assume 1 meter from the ground to the hip and that there is no resistance in the fall.

Using conservation of energy, the speed at impact is
v=sqrt(2*10*1)
v=4.472 m/s

In order to reduce the speed to 1.3 m/s in 2.0 cm, the following equation must be satisfied:

1.3=4.472-(a+9.81)*t
and
2.0/100=4.472*t-.5*(a+9.81)*t^2

the t is the time of compression and a is the deceleration.

The easiest way to solve is to isolate a*t from the first equation and substitute into the second


4.472-1.3+9.81*t=a*t

1/25=2*4.472*t-(4.472-1.3-+9.81*t)*t-9.81*t^2
solve the quadratic for t
and then solve for a
j

Answer 2

Hip Pads For Elderly

Answer 3

s = v0t + (1/2)at^2
v = v0 + at and t = (v - v0)/a
s = v0(v - v0)/a + (1/2)a (v - v0)^2/a^2
2as = 2vv0 - 2v0^2 + (v^2 - 2vv0 + v0^2)
2as = v^2 - v0^2
v^2 = v0^2 + 2as

v0 = 2
v = 1.3
s = 2 cm = .02 m

1.3^2 = 2^2 + 2a(.02)
a = -57.75 m/s^2 = 5.893 g's

v = v0 + at = 2 - 57.75t = 1.3
t = .01212 seconds