Question

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 and (b) deceleration. Express each in multiples of g (9.80 m/s2) by taking its ratio to the acceleration of gravity.

Answer 1

Step-by-step explanation:

Given

Initial velocity(u)=0

Final Velocity(v)=282 m/s

time taken=5 s

acceleration to acquire 282 m/s

`a=(v-u)/(t)`

`a=(282-0)/(5)`

`a=56.4 m/s^2`

a=5.74g

(b)deceleration

It comes to rest in 1.4 s

initial velocity of Just before deceleration is 282 m/s

v=u+at

`0=282-a* 1.4`

`a=(282)/(1.4)=201.42 m/s^2`

a in terms of g

`(201.42)/(9.81)=20.53g`