Question

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?

Answer 1

(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

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

(c) 0.28 s

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:

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