General Physics Please explain thoroughly. Thank you! 2) A human cannot withstan
ID: 1284057 • Letter: G
Question
General Physics
Please explain thoroughly. Thank you!
2) A human cannot withstand a deceleration greater than 820.0 ft/s2 without physical damage occurring. If the deceleration is greater, their body may come to a halt, but their organs inside will keep traveling at high velocity, slamming into their skull and rib cage, causing physical damage or death. Supposing a car crashes into a brick wall at 35.0 mph. We'll model a car airbag as a cylinder with constant cross-sectional diameter, D, but having a variable thickness, r, as shown in Fig.-2a. We'll also assume that there's no friction anywhere, that the car doesn't crumple upon hitting the wall, and that all motion that occurs during a crash, including the expansion of the airbag, is purely horizontal, as shown in Figs.-2b through 2d. Thus, when the car stops moving upon hitting the wall, the driver keeps moving at 35.0 mph (due to Newton's First Law). Assume that the distance of the driver's face from the undeployed airbag is L = 18.0 in, as shown in Fig.-2b, and that the fully inflated airbag has a thickness of rmax = 14.0 in. a. How long does it take for the airbag to fully inflate if it has to be fully inflated by the time the driver's face makes contact with it, as shown in Fig.-2c? (Assume that the airbag inflates at a constant rate.) b. How fast (in mph) will the front surface of the airbag be moving during this time? c. How long will it take for the airbag to fully deflate, as shown in Fig.-2d, if the driver is to survive the crash?Explanation / Answer
a) The driver continues to move at a speed of 35mph even after the car crashes. The drivers face is 18 inches from the uninflated airbag. But since the fully inflated airbag has a thickness of 14 inches, hence the drivers face would move 18-14=4 inches before the airbag inflates to meet him. This distance of 4 inches or (1/3 foot) is covered at a speed of 35mph=35X5280/3600 ft/sec Hence the time take by the airbag to fully inflate= 0.0065 seconds
b) The airbag covers a distance of 14 inches during the time taken to meet the drivers face. Hence the speed of the front surface of the airbag would be 14 inches (or 1.16667 feet/0.0065 seconds = 179.487 ft/sec
c) We know the initial speed of the drivers face = 35mph= 35X5280/3600 ft/sec = 51.3333 ft/sec
We also know that the deceleration provided by the airbag is 820 ft/s2
We have the equation by Newton's laws of motion:
v=u+at where v=final velocity=0, u=initial velocity=51.333ft/s and a=-820ft/s2
Hence, we have t=51.333/820=0.0626 seconds
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