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.

a) 1. 15.23 g, 15.23 g
b) 2. 20.23 g, 20.23 g
c) 3. 20.23 g, 15.23 g
d) 4. 15.23 g, 20.23 g

Answer 1

Dr. John Paul Stapp's acceleration in a rocket sled was 56.4 m/s², equivalent to approximately 5.76 g, and his deceleration was -201.43 m/s², equivalent to approximately 20.55 g.

Step-by-step explanation:

The problem describes Dr. John Paul Stapp's experiences when he rode a rocket sled and requires calculating his acceleration and deceleration. To find the acceleration, we use the formula a = Δv / Δt, where Δv is the change in velocity and Δt is the change in time. Stapp's sled went from rest to 282 m/s in 5.00 s, so the acceleration is (282 m/s - 0 m/s) / 5.00 s = 56.4 m/s². To express this acceleration in multiples of g (9.80 m/s²), we take the ratio of 56.4 m/s² to 9.80 m/s², resulting in about 5.76 g. For the deceleration, Stapp returned to rest in 1.40 s, therefore the deceleration is (0 m/s - 282 m/s) / 1.40 s = -201.43 m/s² (negative sign indicates deceleration). The magnitude of this deceleration in multiples of g is 201.43 m/s² / 9.80 m/s², which gives about 20.55 g.