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3 years ago

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Dr. John Paul Stapp was U.S. Air Force officer whostudied the effects of extreme deceleration on thehuman body. On December 10,

1954, Stapp rode a rocketsled, accelerating from rest to a top speed of 282 m/s(1015 km/h) in 5.00 s, and was brought jarringly back torest in only 1.40 s! Calculate his:
a. Acceleration
b. Deceleration.

Express each in multiples ofby taking its ratio to the acceleration ofgravity.

1 answer:

Ira Lisetskai [31]3 years ago

3 0

Answer:

(a) a = 56.4 m/s², his acceleration a, in multiples of gravity g, is 5.76 g

(b) a = -201.43 m/s², his deceleration -a, in multiples of gravity g, is -20.56 g

Explanation:

(a)

When moving upwards, the initial velocity, u = 0 (he accelerated from rest)

When moving upwards, the final velocity, v = 282 m/s

time of motion during this acceleration, t = 5 s

His acceleration is calculated as;

v = u + at

282 = 0 + 5a

a = 282 / 5

a = 56.4 m/s²

Ratio of his acceleration, a to gravity, g = a/g = 56.4 / 9.8 = 5.76

a = 5.76 g

(b)

When moving downwards, the initial velocity, u = 282 m/s

When moving downwards, the final velocity, v = 0 (he was brought to rest)

time of motion during this deceleration, t = 1.4 s

His deceleration is calculated as;

v = u + at

0 = 282 + 1.4a

1.4a = -282

a = -282 / 1.4

a = -201.43 m/s²

Ratio of his deceleration, -a to gravity, g = -a/g = 201.43 / 9.8 = 20.56

a = -20.56 g

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<u>Knwons</u>

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