Physics, 20.09.2019 18:00 jdjdjdjdjjffi7273

Colonel john p. stapp, usaf, participated in studying whether a jet pilot could survive emergency ejection. on march 19, 1954, he rode a rocket-propelled sled that moved down a track at a speed of 632 mi/h. he and the sled were safely brought to rest in 1.40 s. col. john stapp and his rocket sled are brought to rest in a very short time interval. stapp's face is contorted by the stress of rapid negative acceleration. (courtesy of u. s. air force) (a) determine the negative acceleration he experienced (in m/s2). m/s2 (b) determine the distance he traveled during this negative acceleration (in m). m (c) what if? col. stapp was able to walk away from this experiment. if the human body can survive a negative acceleration five times that experienced by col. stapp, what minimum stopping time (in s) would this correspond to in the 1954 experiment?

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Answer from: Samaritan

(a) -201.8 m/s^2

First of all, let's convert the initial velocity from miles/hour to m/s:

$u=632 mi/h \cdot \frac{1609 m/mi}{3600 s/h}=282.5 m/s$

Then we can find the acceleration by using the equation:

$a=\frac{v-u}{t}$

where:

v = 0 is the final velocity

u = 282.5 m/s is the initial velocity

t = 1.40 s is the time interval

Substituting into the equation,

$a=\frac{0-282.5}{1.40}=-201.8 m/s^2$

(b) 197.7 m

We can find the stopping distance by using the SUVAT equation:

v^2-u^2=2ad

where

v = 0 is the final velocity

u = 282.5 m/s is the initial velocity

a=-201.8 m/s^2 is the acceleration

d is the distance travelled while stopping

Solving for d, we find

$d=\frac{v^2-u^2}{2a}=\frac{0^2-(282.5)^2}{2(-201.8)}=197.7 m$

In this case, the acceleration is five time that experienced in the previous experiment:

a=5(-201.8 m/s^2)=-1009 m/s^2

We can find the stopping time by using the equation:

$a=\frac{v-u}{t}$

where

v = 0 is the final velocity

u = 282.5 m/s is the initial velocity

Since we have the acceleration, we can re-arrange the formula to find the time:

$t=\frac{v-u}{a}=\frac{0-282.5}{-1009}=0.28 s$

Answer from: Quest

b. latent learning (the word "latent" itself means something that exists but isn't obvious, only under certain conditions! )

Answer from: Quest

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