After World War II the U.S. Air Force carried out experiments on the amount of acceleration a human can survive.These experiments led by Jon Stapp, were the first to use crash dummies as well as human subjects, especially Stapp himself, who became an effective advocate for automobile safety belts. In one of the experiments Stapp rode a rocket sled that decelerated from $\mathbf{140}\mathit{m}\mathbf{/}\mathit{s}$(about $\mathbf{310}\mathbf{}\mathit{m}\mathit{i}\mathit{l}\mathbf{/}\mathit{h}$) to $\mathbf{70}\mathit{m}\mathbf{/}\mathit{s}$in just $\mathbf{0}\mathbf{.}\mathbf{6}\mathit{s}$. What was the absolute value of the (negative) average acceleration? (b) The acceleration of a falling object if air resistance is negligible is $\mathbf{9}\mathbf{.}\mathbf{8}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}\mathbf{/}\mathit{s}$,called “one g.” What was the absolute value of the average acceleration in gas? What was the absolute value of the average acceleration in g’s? (Stapp eventually survived a test at $\mathbf{46}\mathbf{}\mathit{g}\mathbf{\text{'}}\mathit{s}\mathbf{}\mathbf{}\mathbf{!}$)

The given data can be listed below as

• The rocket has decelerated from the speed of$\mathbf{140}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$.

• The rocket has decelerated to a speed of$\mathbf{70}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$.

• The rocket has decelerated in about$\mathbf{0}\mathbf{.}\mathbf{6}\mathbf{}\mathit{s}$.

This law illustrates that the acceleration of an object produced by a net force is directly proportional to the magnitude of the net force.

The equation of the average acceleration gives the absolute value of the average acceleration.

1. From Newton’s second law, the equation of the average acceleration can be expressed as:

$a_{avgacc}=\frac{v_2-v_1}{t_2-t_1}$

Here, the decelerated velocity $v_2$is $\mathbf{70}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$and is the initial velocity $v_1$of the rocket is $\mathbf{140}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$, and decelerated time $t_2$is $\mathbf{0}\mathbf{.}\mathbf{6}\mathit{s}$and is the initial time of the rocket $t_1$is $0s$.

For $v_{1,}{v}_{2,}{t}_2$, and $t_1$acceleration is calculated as:

$$\begin{gathered} {\mathit{a}}_{\mathbf{a}\mathbf{v}\mathbf{g}\mathbf{}\mathbf{a}\mathbf{c}\mathbf{c}}\mathbf{}\mathbf{=}\frac{\mathbf{70}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{-}\mathbf{140}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}}{\mathbf{0}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{s}\mathbf{-}\mathbf{0}\mathbf{}\mathbf{s}} \\ {\mathit{a}}_{\mathbf{a}\mathbf{v}\mathbf{g}\mathbf{}\mathbf{a}\mathbf{c}\mathbf{c}}\mathbf{}\mathbf{=}\mathbf{-}\mathbf{116}\mathbf{.}\mathbf{67}\mathbf{.}\left(\frac{1m/s}{1s}\right) \\ {\mathit{a}}_{\mathbf{a}\mathbf{v}\mathbf{g}\mathbf{}\mathbf{a}\mathbf{c}\mathbf{c}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\mathbf{116}\mathbf{.}\mathbf{67}\mathbf{.}\left(1m/s^2\right) \\ {\mathit{a}}_{\mathbf{a}\mathbf{v}\mathbf{g}\mathbf{}\mathbf{a}\mathbf{c}\mathbf{c}}\mathbf{}\mathbf{=}\mathbf{-}\mathbf{116}\mathbf{.}\mathbf{67}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}} \end{gathered}$$

Thus, the absolute value of the (negative) average acceleration is$-116.67m/s^2$.

1. As $1g=9.8m/s^2$, the average acceleration in terms of g is$$\begin{gathered} a_{avgacc}=\frac{-116.67m/s^2}{9.8m/s^2}\times 9.8m/s^2 \\ {a}_{avgacc}=-11.9g \end{gathered}$$

Thus, average acceleration in terms of gravitational acceleration is written as $-11.9g$.