BIO Prevention of Hip Fractures.? 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.
Solution 26E Step 1: We can use Newton’s equations of motion to solve this problem. a) Here, they have provided that, the initial velocity during the impact is, u = 2 m/s We have to reduce it up to 1.3 m/s during the impact in order to prevent the fracture. Therefore, the final velocity during impact should be, v = 1.3 m/s The initial size of the pad is, t = 5 cm = 0.05 m 1 The size of the pad during compression, t = 2 cm = 0.02 m 2 2 2 Therefore, we can use the equation, v = u + 2aS Here, S = (t - t ) 1hich is2othing but the change in thickness of the pad during the compression. Putting all those values in the equation, we get, 2 2 2 2 2 2 1.3 m /s = 2 m /s + (2 × a × [0.05 m - 0.02 m]) 1.69 m /s = 4 m /s + (0.06 m × a ) (1.69 m /s - 4 m /s ) = (0.06 m × a ) 2 2 - 2.31 m /s = 0.06 a m 2 2 2 Therefore, a = - 2.31 m /s / 0.06 m = - 38.50 m/s This is the constant acceleration needed for reducing the velocity from 2 m/s to 1.3 m/s if the pad is compressed from 5 cm thickness to 2 cm thickness. 2 We know that, g = 9.8 m/s Therefore, a = - 38.50 m/s / 9.8 m/s = - 3.928 g