/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 70 Prevention of hip fractures. Fal... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(\mathrm{m} / \mathrm{s}\) . If this can be reduced to 1.3 \(\mathrm{m} / \mathrm{s}\) 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 \(\mathrm{cm}\) thick and compresses by 2.0 \(\mathrm{cm}\) during the impact of a fall, what acceleration (in \(\mathrm{m} / \mathrm{s}^{2}\) and in \(g^{\prime}\) s) does the hip undergo to reduce its speed to 1.3 \(\mathrm{m} / \mathrm{s} ?\) (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.

Short Answer

Expert verified
The acceleration is -65.25 m/s² or -6.65 g. The impact lasts about 0.011 seconds.

Step by step solution

01

Identify Variables and Required Equations

We start by identifying the variables given. Initial speed, \( v_i = 2.0 \, \mathrm{m/s} \); final speed, \( v_f = 1.3 \, \mathrm{m/s} \); pad compresses, \( 2.0 \, \mathrm{cm} = 0.02 \, \mathrm{m} \). We need to find acceleration \( a \), using the equation: \( v_f^2 = v_i^2 + 2a d \), where \( d \) is the distance over which the impact is absorbed.
02

Rearrange and Solve for Acceleration

Rearrange the equation to find acceleration: \( a = \frac{v_f^2 - v_i^2}{2d} \). Substitute the known values into the equation: \( a = \frac{(1.3 \, \mathrm{m/s})^2 - (2.0 \, \mathrm{m/s})^2}{2 \times 0.02 \, \mathrm{m}} \). Calculate the acceleration.
03

Calculate Acceleration in \(g\)

Convert acceleration from \( \mathrm{m/s}^2 \) to \( g \) by dividing by \( 9.81 \, \mathrm{m/s}^2 \). This will express the acceleration as a multiple of gravitational acceleration.
04

Calculate Time of Impact

Using the formula \( v_f = v_i + at \), rearrange to solve for time \( t \): \( t = \frac{v_f - v_i}{a} \). Substitute the values of \( v_i \), \( v_f \), and \( a \) they were calculated previously.
05

Final Calculation

Substitute all values and calculate each required answer for acceleration and time.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinematics
Kinematics is a fundamental concept in physics that deals with the motion of objects without considering the causes of this motion, such as forces. It includes concepts such as displacement, velocity, and acceleration. In the context of hip protection during falls, kinematics helps us understand how quickly the hip changes its velocity upon impact.
To maintain body safety, it's imperative to manage the velocity alteration within safe limits. By finding the initial and final speeds, we can measure how velocity changes over a compressed distance when an elastic hip pad is used. This introduces another key concept—acceleration, crucial for our calculations.
Acceleration
Acceleration refers to the rate at which an object's velocity changes over time. In formulas, we often denote it by the letter \( a \). In physics, it's typically measured in meters per second squared \(\mathrm{m/s^2}\). For the hip impact case, we're calculating how much the hip's speed decreases due to the pad's cushioning effect.
First, we understand that the hip's velocity needs to drop from 2.0 \(\mathrm{m/s}\) to 1.3 \(\mathrm{m/s}\). This velocity change occurs over the 0.02 meters of pad compression. Utilizing the kinematic equation \( v_f^2 = v_i^2 + 2ad \), we rearrange it to find acceleration: \( a = \frac{v_f^2 - v_i^2}{2d} \). This acceleration tells us how rapidly the velocity changes at impact. Converting this into \( g \) units (gravitational accelerations) helps us understand in more relatable terms, as 1 \( g \) is the normal acceleration due to gravity.
Elastic collision
An elastic collision occurs when two objects collide and bounce apart without losing energy to deformation or heat. While the hip pad isn't involved in a perfectly elastic collision (since energy is absorbed during compression), it does serve to lessen the impact force exerted on the hip, allowing for energy absorption and reducing the velocity upon impact.
Enhancing safety gear such as hip pads leverages elastic behavior—meaning they compress under force to absorb part of the energy. By doing so, they reduce the effective velocity at which a person hits the ground, thus minimizing the chance of fracture. These pads are designed to exhibit elastic properties, carefully managing energy more efficiently.
Impact force
The concept of impact force is crucial when considering safety equipment like hip pads. These forces occur when objects collide and are crucial to understand in reducing injury likelihood. Despite the cushion's compression and controlled elasticity aiding in energy absorption, the force exerted on the hip remains substantial if not well-managed.
The primary goal of wearing hip pads is to lessen the force of impact. By helping manage the energy exchange efficiently during the short duration of the force application, the hip pads enable the impact to spread out rather than peaking sharply. Utilizing impact force formulas aids in calculating the forces experienced by the hip during a fall, balancing force, time, and protective measures to prevent fractures.

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Most popular questions from this chapter

A test driver at Incredible Motors, Inc., is testing a new model car having a speedometer calibrated to read m/s rather than mi/h. The following series of speedometer readings was obtained during a test run: $$\begin{array}{llllllllll}{\text { Time (s) }} & {0} & {2} & {4} & {6} & {8} & {10} & {12} & {14} & {16} \\ {\text { Velocity }(\mathrm{m} / \mathrm{s})} & {0} & {0} & {2} & {5} & {10} & {15} & {20} & {22} & {22}\end{array}$$ (a) Compute the average acceleration during each 2 s interval. Is the acceleration constant? Is it constant during any part of the test run? (b) Make a velocity-time graph of the data shown, using scales of \(1 \mathrm{cm}=1\) s horizontally and \(1 \mathrm{cm}=\) 2 \(\mathrm{m} / \mathrm{s}\) vertically. Draw a smooth curve through the plotted points. By measuring the slope of your curve, find the magnitude of the instantaneous acceleration at times \(t=9 \mathrm{s}, 13 \mathrm{s}\) and 15 \(\mathrm{s}\) .

Sound travels at a speed of about 344 \(\mathrm{m} / \mathrm{s}\) in air. You see a distant flash of lightning and hear the thunder arrive 7.5 seconds later. How many miles away was the lightning strike? (Assume the light takes essentially no time to reach you.)

How high is the cliff? Suppose you are climbing in the High Sierra when you suddenly find yourself at the edge of a fog-shrouded cliff. To find the height of this cliff, you drop a rock from the top and, 10.0 s later, hear the sound of it hitting the ground at the foot of the cliff. (a) Ignoring air resistance, how high is the cliff if the speed of sound is 330 \(\mathrm{m} / \mathrm{s} ?\) (b) Suppose you had ignored the time it takes the sound to reach you. In that case, would you have overestimated or underestimated the height of the cliff? Explain your reasoning.

A car driving on the turnpike accelerates uniformly in a straight line from 88 \(\mathrm{ft} / \mathrm{s}\) (60 mph) to 110 \(\mathrm{ft} / \mathrm{s}\) (75 mph) in 3.50 \(\mathrm{s} .\) (a) What is the car's acceleration? (b) How far does the car travel while it accelerates?

Freeway ramps. In designing a freeway on-ramp, you must make it long enough for cars to be able to accelerate and merge safely with traffic traveling at speeds of 70 mph. For an off-ramp, the cars must be able to come to a stop from 70 mph over the length of the ramp. Your traffic engineers provide the following data: Powerful cars can accelerate from rest to 60 mph in 5.0 s, while less powerful cars take twice as long to do this. When slowing down from 60 mph, a car with good tire treads will take 12.0 s to stop, whereas one with bald tires takes 20.0 s. In order to safely accommodate all the types of vehicles, what should be the minimum length (in meters) of on-ramps and off- ramps? Assume constant acceleration.

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