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Answer :
A. [tex]2249.25 m/s^2[/tex] is the magnitude of the acceleration.
B. An average force of approximately 179,940 N is applied to the strap from the restraint system.
a) The equation of motion can be used to calculate the magnitude of acceleration:
[tex]v^2 = u^2 + 2as[/tex]
Where:
v = final velocity = 67 m/s
u = initial velocity = 0 m/s
s = displacement = 360 m
When we rewrite the equation, we get:
[tex]a = (v^2 - u^2) / (2s)\\a = (67^2 - 0^2) / (2 * 360)[/tex]
a = 2249.25 m/s²
As a result, [tex]2249.25 m/s^2[/tex] is the magnitude of the acceleration.
b) We can apply Newton's second law of motion to obtain the average force exerted by the restraining system:
F = ma
Where:
m = mass = 80 kg
a = acceleration (from part A) = 2249.25 m/s²
F = 80 *2249.25
F = 179,940 N
Therefore, an average force of approximately 179,940 N is applied to the strap from the restraint system.
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Your question is incomplete, most probably the complete question is:
Col. John Stapp crash tests From 1946 through 1958, Col. John Stapp headed the U.S. Air Force Aero Medical Laboratory's studies of the human body's ability to tolerate high accelerations during plane crashes. Conventional wisdom at the time indicated that a plane's negative acceleration should not exceed 180 m/s² (18 times gravitational acceleration, or 18g). Stapp and his colleagues built a 700-kg “Gee Whiz” rocket sled, track, and stopping pistons to measure human tolerance to high acceleration. Starting in June 1949, Stapp and other live subjects rode the sled. In one of Stapp's rides, the sled started at rest and 360 m later was traveling at speed 67 m/s when its braking system was applied, stopping the sled in 6.0 m. He had demonstrated that 18g was not a limit for human deceleration.
A) What is the magnitude of the acceleration of Stapp and his sled as their speed increased from zero to 67 m/s?
B) What is the average force exerted by the restraining system on 80-kg Stapp while his speed decreased from 67 m/s to zero in a distance of 6.0 m?
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