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Dr. John Paul Stapp was a U.S. Air Force officer who studied the effects of extreme deceleration on the human body. On December 10, 1954, Stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.00 s, and was brought jarringly back to rest in only 1.40 s! Calculate his acceleration and deceleration. Express each in multiples of g (9.80 m/s2) by taking its ratio to the acceleration of gravity.

Respuesta :

Answer:

(a) Acceleration = 5.755 g

(b) Deceleration = 20.55 g

Explanation:

We have given that rocket accelerates from rest to 282 m/sec

So initial velocity u = 0 m/sec

And final velocity v = 282 m/sec

And time t = 5 sec

From first equation of motion v = u+at

So acceleration [tex]a=\frac{v-u}{t}=\frac{282-0}{5}=56.4 m/sec^2[/tex]

We know that g = 9.8 [tex]m/sec^2[/tex]

So 56.4  [tex]m/sec^2[/tex] = [tex]=\frac{56.4}{9.8}=5.755g[/tex]

(b) Now initial velocity u = 282 m/sec

And finally it comes to rest so final velocity v = 0 m/sec

Time t = 1.4 sec

So acceleration a [tex]=\frac{v-u}{t}=\frac{0-282}{1.4}=-201.428m/sec^2[/tex]

So deceleration = 201.428 [tex]m/sec^2[/tex]

We know that g = 9.8 [tex]m/sec^2[/tex]

So 201.428  [tex]m/sec^2[/tex] = [tex]=\frac{201.428}{9.8}=20.55g[/tex]

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