91影视

Velocity of the spacecraft is $<-20,-90,40>\mathrm{m}/\mathrm{s}$.

Initial position of the spacecraft is $<200,300,-500>$.

Final position of the spacecraft is $<-380,-2310,660>m$.

The position vector of a location $<x_1,y_1,z_1>$ with respect to another location $<x_2,y_2,z_2>$is

$\stackrel{\mathbf{鈫�}}{\mathbf{r}}\mathbf{=}\left(x_2-x_1\right)\hat{\mathbf{i}}\mathbf{+}\left(y_2-y_1\right)\hat{\mathbf{j}}\mathbf{+}\left(z_2-z_1\right)\hat{\mathbf{k}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}.....\left(\mathrm{l}\right)$

The magnitude of a vector$\stackrel{鈫�}{r}=x\hat{i}+y\hat{j}+z\hat{k}$is

The unit vector along a vector $\stackrel{鈫�}{r}$is

$\hat{\mathbf{r}}\mathbf{=}\frac{\stackrel{\mathbf{鈫�}}{\mathbf{r}}}{\left|\stackrel{鈫�}{r}\right|}\mathbf{}.....\left(\mathrm{lll}\right)$

From equation (I), the velocity vector of the spacecraft with respect to the origin $<0,0,0>$is

$\stackrel{鈫�}{v}=\left(-20\hat{i}-90\hat{j}+40\hat{k}\right)\mathrm{m}/\mathrm{s}$

Displacement vector is the position vector of the final position with respect to the initial position. From equation (I), the displacement vector of the spacecraft is

$$\begin{gathered} \stackrel{鈫�}{r}=\left\{\left(200-\left(-380\right)\right)\hat{i}+\left(300-\left(-2310\right)\right)\hat{j}+\left(-500-660\right)\hat{k}\right\}\mathrm{m} \\ =\left(580\hat{i}+2610\hat{j}-1160\hat{k}\right)\mathrm{m} \end{gathered}$$

The time of travel is obtained by dividing the magnitude of position by magnitude of velocity along any one coordinate. The operation along the other coordinates will yield the same result. Thus, the operation along x axis yields

$$\begin{gathered} t=\frac{580\mathrm{m}}{20\mathrm{m}/\mathrm{s}} \\ =29\mathrm{s} \end{gathered}$$

Thus, the required time is $29\mathrm{s}$.

Distance traveled is the magnitude of displacement vector. From equation (II), the magnitude of $\stackrel{鈫�}{r}$ is

$$\begin{gathered} \left|\stackrel{鈫�}{r}\right|=\sqrt{{\left(580\right)}^2+{\left(2610\right)}^2+{\left(-1160\right)}^2}\mathrm{m} \\ =\sqrt{336400+6812100+1345600}\mathrm{m} \\ =2914.46\mathrm{m} \end{gathered}$$

Thus, the required distance is 2914.46 m .

Speed is the magnitude of velocity vector. From equation (II), the magnitude of $\stackrel{鈫�}{v}$is

$$\begin{gathered} \left|\stackrel{鈫�}{v}\right|=\sqrt{{\left(-20\right)}^2+{\left(-90\right)}^2+{\left(40\right)}^2}\mathrm{m}/\mathrm{s} \\ =\sqrt{400+8100+1600}\mathrm{m}/\mathrm{s} \\ =100.5\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the required speed is 100.5 m/s .

From equation (III), the unit vector along $\stackrel{鈫�}{v}$ is

$$\begin{gathered} \hat{v}=\frac{\left(-20\hat{\mathrm{i}}-90\hat{\mathrm{j}}+40\hat{\mathrm{k}}\right)\mathrm{m}/\mathrm{s}}{100.5\mathrm{m}/\mathrm{s}} \\ =-0.199\hat{i}-0.896\hat{j}+0.398\hat{k} \end{gathered}$$

Thus, the required unit vector is $-0.199\hat{i}-0.896\hat{j}+0.398\hat{k}$.