Question

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 5.00 s and was brought jarringly back to rest in only 1.40 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.

Answer 1

56.4 m/s², a = 5.75510204082 g

-201.428571429 m/s², a = -20.5329838358 g

Step-by-step explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.8 m/s²

`v=u+at\\\Rightarrow a=(v-u)/(t)\\\Rightarrow a=(282-0)/(5)\\\Rightarrow a=56.4\ m/s^2`

The acceleration is 56.4 m/s²

Dividing by g

`(a)/(g)=(56.4)/(9.8)\\\Rightarrow a=5.75510204082g`

a = 5.75510204082 g

`v=u+at\\\Rightarrow a=(v-u)/(t)\\\Rightarrow a=(0-282)/(1.4)\\\Rightarrow a=-201.428571429\ m/s^2`

The acceleration is -201.428571429 m/s²

Dividing by g

`(a)/(g)=(-201.428571429)/(9.8)\\\Rightarrow a=-20.5539358601g`

a = -20.5329838358 g