Question

A spacecraft traveling at a velocity of $\mathbf{<}\mathbf{-}\mathbf{20}\mathbf{,}\mathbf{-}\mathbf{90}\mathbf{,}\mathbf{40}\mathbf{>}\mathbf{m}\mathbf{/}\mathbf{s}$is observed to be at a location $\mathbf{<}\mathbf{200}\mathbf{,}\mathbf{300}\mathbf{,}\mathbf{-}\mathbf{500}\mathbf{>}$relative to an origin located on a nearby asteroid. At a later time the spacecraft is at location $\mathbf{<}\mathbf{-}\mathbf{380}\mathbf{,}\mathbf{-}\mathbf{2310}\mathbf{,}\mathbf{660}\mathbf{>}\mathbf{m}$.

(a)How long did it take the spacecraft to travel between these locations?

(b)How far did the spacecraft travel?

(c)What is the speed of the spacecraft?

(d)What is the unit vector in the direction of the spacecraft’s velocity?

Short answer

(a) It took 29 s for the spacecraft to travel between the locations.

(b) The spacecraft traveled 2914.46 m.

(c) The speed of the spacecraft is 100.5 m/s.

(d) The unit vector in the direction of the spacecraft's velocity is $\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{199}\hat{\mathbf{i}}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{896}\hat{\mathbf{j}}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{398}\hat{\mathbf{k}}$.

Step-by-step solution

Given data

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 $<-380,-2310,660>m$.

Position vector, magnitude of a vector and unit vector

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

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

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

$\left|\overrightarrow{r}\right|\mathbf{=}\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}}\mathbf{}\mathbf{}\mathbf{}....\left(\mathrm{ll}\right)$

The unit vector along a vector $\overrightarrow{r}$is

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

Determining the time of travel of the spacecraft

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

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

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

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

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 $29\mathrm{s}$.

Determining the distance traveled by the spacecraft

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

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

Thus, the required distance is 2914.46 m .

Determining the speed of the spacecraft

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

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

Thus, the required speed is 100.5 m/s .

Determining the unit vector along the velocity

From equation (III), the unit vector along $\overrightarrow{v}$ 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 $-0.199\hat{i}-0.896\hat{j}+0.398\hat{k}$.