91Ó°ÊÓ

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

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

${\stackrel{→}{r}}_{\text{position}}=\stackrel{→}{r}+{\stackrel{→}{r}}_1\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}……\text{(I)}$

Here, ${\stackrel{→}{r}}_{\text{position}}$is the final position of the spaceship after engine is shuts off,$\stackrel{→}{r}$is the initial position of the spaceship, and ${\stackrel{→}{r}}_1$is the position of the spaceship after acceleration.

The displacement vector of the spaceship during the acceleration is, ${\stackrel{→}{r}}_1=\stackrel{→}{v}t+\frac{1}{2}\stackrel{→}{a}{t}^2$.

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

Substitute $9500\widehat{i}\text{m}/\text{s}$for $35\text{min}$for $t$and $(40\widehat{i}−20\widehat{k})\text{m}/{\text{s}}^2$for $\stackrel{→}{a}$in the above equation to find${\stackrel{→}{r}}_1$.

$\begin{array}{l}{\stackrel{→}{r}}_1=(9500\widehat{i}\text{m}/\text{s})\left(35\text{min}×\frac{60\text{s}}{1\text{min}}\right)+\frac{1}{2}\left((40\widehat{i}−20\widehat{k})\text{m}/{\text{s}}^2\right){\left(35\text{min}×\frac{60\text{s}}{1\text{min}}\right)}^2\\ =\left(19950×{10}^3\widehat{i}\right)\text{m}+\left(88200×{10}^3\widehat{i}−44100×{10}^3\widehat{k}\right)\text{m}\\ =\left(\left(108.15×{10}^6\widehat{i}\right)\text{m}×\frac{1\text{km}}{{10}^3\text{m}}\right)−\left(\left(44.1×{10}^6\widehat{k}\right)\text{m}×\frac{1\text{km}}{{10}^3\text{m}}\right)\\ =(108.15\widehat{i}−44.1\widehat{k})×{10}^3\text{km}\end{array}$

$\begin{array}{l}{\stackrel{→}{r}}_1=(9500\widehat{i}\text{m}/\text{s})\left(35\text{min}×\frac{60\text{s}}{1\text{min}}\right)+\frac{1}{2}\left((40\widehat{i}−20\widehat{k})\text{m}/{\text{s}}^2\right){\left(35\text{min}×\frac{60\text{s}}{1\text{min}}\right)}^2\\ =\left(19950×{10}^3\widehat{i}\right)\text{m}+\left(88200×{10}^3\widehat{i}−44100×{10}^3\widehat{k}\right)\text{m}\\ =\left(\left(108.15×{10}^6\widehat{i}\right)\text{m}×\frac{1\text{km}}{{10}^3\text{m}}\right)−\left(\left(44.1×{10}^6\widehat{k}\right)\text{m}×\frac{1\text{km}}{{10}^3\text{m}}\right)\\ =(108.15\widehat{i}−44.1\widehat{k})×{10}^3\text{km}\end{array}$

Thus, the position vector of the spaceship after acceleration is $(108.15\widehat{i}−44.1\widehat{k})×{10}^3\text{km}$.

Substitute$(108.15\widehat{i}−44.1\widehat{k})×{10}^3\text{km}$for${\stackrel{→}{r}}_1$and $(600\widehat{i}−400\widehat{j}+200\widehat{k})×{10}^3\text{km}$for $\stackrel{→}{r}$in the equation (I) to find$\stackrel{→}{r}$ position.

Therefore, the position of the spaceship when the engine of the ship was shuts off is

$(708.15\widehat{i}−400\widehat{j}+155.9\widehat{k})×{10}^3\text{km.}$