Chapter 4: Q. 44 (page 107)

A spaceship maneuvering near Planet Zeta is located at $\hat{r} = (600 \hat{i} - 400 \hat{j} + 200 \hat{k}) \times 10^{3} km$ , relative to the planet, and traveling at $\vec{v} = 9500 \hat{ı} m / s$ . It turns on its thruster engine and accelerates with $\vec{a} = (40 \hat{ı} - 20 \hat{k}) m / s^{2}$ for $35 min$ . What is the spaceship's position when the engine shuts off? Give your answer as a position vector measured in $km$ .

Short Answer

Expert verified

The position of the spaceship when the engine of the ship was shuts off is $(708.15 \hat{i} - 400 \hat{j} + 155.9 \hat{k}) \times 10^{3} km.$

Step by step solution

Step 1. Given information

The initial position vector of the spaceship relative to the planet Zeta is $(600 \hat{i} - 400 \hat{j} + 200 \hat{k}) \times 10^{3} km$ , the initial speed of the spaceship is $9500 \hat{i} m / s$ , then the spaceship turns on thruster engine and accelerate with $(40 \hat{i} - 20 \hat{k}) m / s^{2}$ for $35 min$ .

The position of the spaceship when the engine is shut off with reference to the planet Zeta is,

${\vec{r}}_{position} = \vec{r} + {\vec{r}}_{1} \dots \dots (I)$

Here, ${\vec{r}}_{position}$ is the final position of the spaceship after engine is shuts off, $\vec{r}$ is the initial position of the spaceship, and ${\vec{r}}_{1}$ is the position of the spaceship after acceleration.

The displacement vector of the spaceship during the acceleration is, ${\vec{r}}_{1} = \vec{v} t + \frac{1}{2} \vec{a} t^{2}$ .

Here, $\vec{v}$ is the initial velocity vector of the spaceship, $t$ is the time period of acceleration, and $\vec{a}$ is the acceleration vector of the spaceship.

Step 2. Explanation

Substitute $9500 \hat{i} m / s$ for $35 min$ for $t$ and $(40 \hat{i} - 20 \hat{k}) m / s^{2}$ for $\vec{a}$ in the above equation to find ${\vec{r}}_{1}$ .

$\begin{array}{l} {\vec{r}}_{1} = (9500 \hat{i} m / s) (35 min \times \frac{60 s}{1 min}) + \frac{1}{2} ((40 \hat{i} - 20 \hat{k}) m / s^{2}) {(35 min \times \frac{60 s}{1 min})}^{2} \\ = (19950 \times 10^{3} \hat{i}) m + (88200 \times 10^{3} \hat{i} - 44100 \times 10^{3} \hat{k}) m \\ = ((108.15 \times 10^{6} \hat{i}) m \times \frac{1 km}{10^{3} m}) - ((44.1 \times 10^{6} \hat{k}) m \times \frac{1 km}{10^{3} m}) \\ = (108.15 \hat{i} - 44.1 \hat{k}) \times 10^{3} km \end{array}$

Thus, the position vector of the spaceship after acceleration is $(108.15 \hat{i} - 44.1 \hat{k}) \times 10^{3} km$ .

Substitute $(108.15 \hat{i} - 44.1 \hat{k}) \times 10^{3} km$ for ${\vec{r}}_{1}$ and $(600 \hat{i} - 400 \hat{j} + 200 \hat{k}) \times 10^{3} km$ for $\vec{r}$ in the equation (I) to find $\vec{r}$ position.