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Dr. John Paul Stapp was a U.S. Air Force officer who studied the effects of extreme acceleration 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 5.1 s and was brought jarringly back to rest in only 1.9 s. Calculate his (a) acceleration in his direction of motion and (b) acceleration opposite to his direction of motion. Express each in multiples of g (9.80 m/s2) by taking its ratio to the acceleration of gravity. Report each answer rounded to one decimal point and report only positive answers (tak

Respuesta :

Answer:

(a) 5.6g

(b) 15.1g

Explanation:

Let us assume:

  • u = initial speed of the rocket
  • v = final speed of the rocket
  • a = acceleration of the rocket
  • t = time interval
  • g = acceleration due to gravity = [tex]9.80\ m/s^2[/tex]

Part (a):

While Dr. John Paul Stapp accelerates the rocket to its top speed, we have

[tex]u = 0 m/s\\v = 282\ m/s\\t = 5.1\ s\\\therefore a = \dfrac{v-u}{t}\\\Rightarrow  a = \dfrac{282-0}{5.1}\\\Rightarrow  a =55.29\\\textrm{Dividing both sides by }g\\\Rightarrow \dfrac{a}{g}=\dfrac{55.29}{9.80}\\\Rightarrow \dfrac{a}{g}=5.6\\\Rightarrow a=5.6g[/tex]

Hence, the acceleration of the rocket in the direction of motion is 5.6g.

Part (b):

While Dr. John Paul Stapp accelerates the rocket from its top speed to rest, we have

[tex]u = 282 m/s\\v = 0\ m/s\\t = 1.9\ s\\\therefore a = \dfrac{v-u}{t}\\\Rightarrow  a = \dfrac{0-282}{1.9}\\\Rightarrow  a =-148.42\\\textrm{Dividing both sides by }g\\\Rightarrow \dfrac{a}{g}=\dfrac{-148.42}{9.80}\\\Rightarrow \dfrac{a}{g}=-15.1\\\Rightarrow a=-15.1g[/tex]

Here, the negative sign represents that the motion of acceleration of the rocket is opposite to the direction of motion.

Hence, the acceleration of the rocket in the direction opposite to that of motion is 15.1g.

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