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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鈥檚 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$$\begin{gathered} a_{avgacc}=\frac{-116.67m/s^2}{9.8m/s^2}脳9.8m/s^2 \\ {a}_{avgacc}=-11.9g \end{gathered}$$

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