Question

1. an astronaut is piloting her spacecraft toward the international space station. to stop the spacecraft, she fires the retro - rockets, which cause the spacecraft to slow down from 20.0 m/s e to 0.0 m/s in 12 s. (a) what is the acceleration of the spacecraft? (b) what is the displacement of the spacecraft when it comes to rest? a) \\( \overrightarrow { v _ { i } } = 20.0 \mathrm { m } / \mathrm { s } \mathrm { e } \\) \\( v _ { f } = 0.0 \mathrm { m } / \mathrm { s } \\) \\( \delta t = 12 \mathrm { s } \\) \\( \overrightarrow { v _ { f } } = \overrightarrow { v _ { i } } + \overrightarrow { a } \delta t \\) \\( \overrightarrow { a } = \frac { \overrightarrow { v _ { f } } - \overrightarrow { v _ { i } } } { \delta t } \\) \\( = \frac { 0 \mathrm { m } / \mathrm { s } - 20.0 \mathrm { m } / \mathrm { s } } { 12 \mathrm { s } } \\)

Explanation:

Step1: Calculate acceleration

Use the formula \(a=\frac{v_f - v_i}{\Delta t}\). Here, \(v_i = 20.0\ m/s\), \(v_f=0\ m/s\), and \(\Delta t = 12\ s\).

$$ a=\frac{0 - 20.0}{12}=\frac{- 20.0}{12}\approx - 1.67\ m/s^{2} $$

The negative sign indicates the acceleration is in the west direction (opposite to the initial east - ward motion).

Step2: Calculate displacement

Use the formula \(d=\frac{v_i + v_f}{2}\times\Delta t\). Substitute \(v_i = 20.0\ m/s\), \(v_f = 0\ m/s\), and \(\Delta t=12\ s\)

$$ d=\frac{20.0 + 0}{2}\times12=10\times12 = 120\ m $$

Answer:

a. The acceleration of the spacecraft is approximately \(-1.67\ m/s^{2}\) (west - ward).
b. The displacement of the spacecraft when it comes to rest is \(120\ m\) (east - ward).