A space probe in outer space has a mass of $\mathbf{111}\mathbf{}\mathbf{\text{kg}}$, and it is traveling at a speed of$\mathbf{29}\mathbf{}\mathbf{\text{m}}\mathbf{/}\mathbf{\text{s}}$. When it is at location$\mathbf{(}\mathbf{445}\mathbf{,}\mathbf{535}\mathbf{,}\mathbf{-}\mathbf{350}\mathbf{)}\mathbf{\text{m}}$it begins firing two booster rockets. The rockets exert constant forces of$\mathbf{(}\mathbf{90}\mathbf{,}\mathbf{150}\mathbf{,}\mathbf{190}\mathbf{)}\mathbf{\text{N}}$and$\mathbf{(}\mathbf{90}\mathbf{,}\mathbf{-}\mathbf{90}\mathbf{,}\mathbf{-}\mathbf{585}\mathbf{)}\mathbf{\text{N}}$, respectively. The rockets fire until the spacecraft reaches location$\mathbf{(}\mathbf{449}\mathbf{,}\mathbf{541}\mathbf{,}\mathbf{-}\mathbf{354}\mathbf{)}\mathbf{\text{m}}$. Now what is its speed? There is negligible mass loss due to the rocket exhaust.

The mass is$\mathrm{m}=111\text{kg}$

The initial speed is$\text{v}=29\text{m}/\text{s}$

The initial position is${\text{d}}_{\text{i}}=(445,535,-350)\text{\hspace{0.17em}m}$

The final position is${\text{d}}_{\text{f}}=(449,541,-354)\text{\hspace{0.17em}m}$

The constant forces exerted by the rockets,$(90,150,190)\text{\hspace{0.17em}N}$

The displacement of the space probe is$(90,-90,-585)\text{\hspace{0.17em}N}$

Find for the displacement of the probe by subtracting the initial position from the final position.

$\overrightarrow{\mathbf{\text{d}}}\mathbf{=}{\mathbf{\text{d}}}_{\mathbf{\text{f}}}\mathbf{-}{\mathbf{\text{d}}}_{\mathbf{\text{i}}}$$\overrightarrow{\mathbf{\text{d}}}\mathbf{=}{\mathbf{\text{d}}}_{\mathbf{\text{f}}}\mathbf{-}{\mathbf{\text{d}}}_{\mathbf{\text{i}}}$

Apply the energy principle where the change in energy of a system or object is equal to the work done and energy input of the surroundings. This can be expressed as,

${\mathbf{\text{E}}}_{\mathbf{\text{f}}}\mathbf{=}{\mathbf{\text{E}}}_{\mathbf{\text{i}}}\mathbf{+}\mathbf{\text{W}}$

The system has two kinds of energy, the kinetic energy which associated with its motion and the rest energy where it has zero speed. The summation of both energies called the particle energy.

Then can be further expressed$\mathbf{(}{\mathbf{\text{K}}}_{\mathbf{\text{f}}}\mathbf{+}{\mathbf{\text{mc}}}^{\mathbf{2}}\mathbf{)}\mathbf{=}\mathbf{(}{\mathbf{\text{K}}}_{\mathbf{\text{i}}}\mathbf{+}{\mathbf{\text{mc}}}^{\mathbf{2}}\mathbf{)}\mathbf{+}\mathbf{\text{W}}$.

The formula above can be broken down into another expression where in the rest energy $\left({\text{mc}}^2\right)$can be cancelled out since there is no change in its rest energy. In this expression, $\text{K}=\frac{1}{2}{\text{mv}}^2$

$\begin{array}{c}\left(\frac{1}{2}{\mathrm{mv}}_{\text{f}}^2+{\mathrm{mv}}^2\right)=\left(\frac{1}{2}{\mathrm{mv}}_{\text{i}}^2+{\mathrm{mv}}^2\right)+\mathrm{W}\\ \frac{1}{2}{\mathrm{mv}}_{\text{f}}^2=\frac{1}{2}{\mathrm{mv}}_{\text{i}}^2+\text{W}\end{array}$

$\mathbf{W}$can be broken down as $({\mathrm{F}}_1\cdot \overrightarrow{\mathrm{d}}+{\mathrm{F}}_2\cdot \overrightarrow{\mathrm{d}})$

$\begin{array}{c}\overrightarrow{d}=d_f-d_i\\ =(449,541,-354)\text{\hspace{0.17em}m}-(445,535,-350)\text{\hspace{0.17em}m}\\ =(4,6,-4)\text{\hspace{0.17em}m}\end{array}$

$\begin{array}{c}{\text{E}}_{\text{f}}={\text{E}}_{\text{i}}+\text{W}\\ \frac{1}{2}{\text{mv}}_{\text{f}}^2=\frac{1}{2}{\text{mv}}_{\text{i}}^2+({\text{F}}_1\cdot \overrightarrow{\text{d}}+{\text{F}}_2\cdot \overrightarrow{\text{d}})\\ \frac{1}{2}(111\text{kg})({\text{v}}_{\text{f}}^2-{(29{\text{ms}}^{-1})}^2)=\left[\right((90,150,195)\text{\hspace{0.17em}N}\cdot (4,6,-4)\text{\hspace{0.17em}m})+((90,-90,-585)\text{\hspace{0.17em}N}\cdot (4,6,-4)\text{\hspace{0.17em}m}\left)\right]\\ 55.5{\text{v}}_{\text{f}}^2-46675.5\text{\hspace{0.17em}J}=2640\text{\hspace{0.17em}J}\\ {\text{v}}_f=\left(\sqrt{\frac{2640+46675.5}{55.5}}\right)\text{\hspace{0.17em}m/s}\\ =29.8\text{\hspace{0.17em}m/s}\end{array}$

Therefore, the speed of the rocket is$29.8\text{m}/\text{s}$ .