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

Answer:

Explanation:

Given

Initial velocity(u)=0

Final Velocity(v)=282 m/s

time taken=5 s

acceleration to acquire 282 m/s

$a=\frac{v-u}{t}$

$a=\frac{282-0}{5}$

$a=56.4 m/s^2$

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

$0=282-a\times 1.4$

$a=\frac{282}{1.4}=201.42 m/s^2$

a in terms of g

$\frac{201.42}{9.81}=20.53g$