Question

As your spaceship coasts towards Mars, you need to move a heavy load of 1200 kg along a hallway of the spacecraft that has a ${\mathbf{90}}^{\mathbf{\circ }}$right turn, without touching the walls, floor, or ceiling, by working remotely, using devices attached to the load that can be programmed to fire blasts of compressed air for up to in any$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\hspace{0.33em}}\mathbf{s}\mathbf{}$ desired direction. During a blast the load is subjected to a force of $\mathbf{}\mathbf{20}\mathbf{\hspace{0.33em}}\mathbf{N}$. The center of the load must move 3 m along the first section of the hallway, starting from rest, then 4 m along the second section, ending at rest. Let the starting point be $\mathbf{⟨}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{⟩}\mathbf{\hspace{0.33em}}\mathbf{m}$, with the first section ending at $\mathbf{⟨}\mathbf{0}\mathbf{,}\mathbf{3}\mathbf{,}\mathbf{0}\mathbf{⟩}\mathbf{\hspace{0.33em}}\mathbf{m}$and the second section ending at $\mathbf{⟨}\mathbf{4}\mathbf{,}\mathbf{3}\mathbf{,}\mathbf{0}\mathbf{⟩}\mathbf{\hspace{0.33em}}\mathbf{m}$. Using just three blasts of compressed air, choose the times when these blasts should be scheduled, their durations, and their directions. How long does it take to complete the entire move?

Short answer

Te blast should be fired at$12.42\hspace{0.33em}s$ in the- y direction, $13.42\hspace{0.33em}s$in the +x direction and $27.91\hspace{0.33em}s$in the -x direction.

The entire move will require$28.91\hspace{0.33em}s$ .

Step-by-step solution

Identification of the given data

The given data can be listed below as:

• The mass of the heavy load is$\mathrm{m}=1200\hspace{0.33em}\mathrm{kg}$ .
• The angle of the load is$\theta ={90}^{\circ }$ .
• The time taken for the blast is $\mathrm{t}=1.0\hspace{0.33em}\mathrm{s}$.
• The force exerted on the blast is $F=20\hspace{0.33em}N$.
• The initial distance moved by the load is$h_1=3\hspace{0.33em}m$ .
• The final distance moved by the load is $h_2=4\hspace{0.33em}m$.
• The starting point of the blast is$s_1=⟨0,0,0⟩\hspace{0.33em}m$ .
• The middle point of the blast is $s_2=⟨0,3,0⟩\hspace{0.33em}m$.
• The ending point of the blast is $s_3=⟨4,3,0⟩\hspace{0.33em}m$.

Significance of the average velocity

The average velocity is described as the division of the change in the displacement with the change in the time. Moreover, the average velocity is the average speed that has a particular direction.

Determination of the times

It has been identified that the blast first occurs in the - y direction, then in the +x direction and finally, in the -x direction.

The equation of the time for scheduling the blast initially is expressed as:

$\mathrm{m}\frac{{\mathrm{s}}_2-{\mathrm{s}}_1}{{\mathrm{t}}_1}={\mathrm{p}}_1+{\mathrm{Ft}}_1$

Here, m is the mass of the heavy load, $s_2$is the middle point of the blast,$s_1$is the starting point of the blast, $p_1$is the initial momentum, F is the force exerted on the blast and $t_1$is the time taken to schedule the blast initially.

As initially the spaceship was at rest, then the initial velocity of the spaceship is zero.

Substitute the values in the above equation.

As one second is needed to produce the required force, then the time needed to fire the blast is 12.42 s .

The equation of the time for scheduling the blast finally is expressed as:

$\mathrm{m}\frac{{\mathrm{s}}_3-{\mathrm{s}}_2}{{\mathrm{t}}_2}={\mathrm{p}}_1+{\mathrm{Ft}}_2$

Here, m is the mass of the heavy load,$s_3$ is the final point of the blast,$s_2$ is the middle point of the blast,$p_1$ is the initial momentum, F is the force exerted on the blast and $t_2$is the time taken to schedule the blast finally.

As initially the spaceship was at rest, then the initial velocity of the spaceship is zero

.Substitute the values in the above equation.

The equation of the total time is expressed as:

$t=t_1+t_2$

Here, t is the total time.

Substitute the values in the above equation.

$$\begin{gathered} t=13.42s+15.49s \\ =28.91s \end{gathered}$$

As one second is needed to produce the required force, then the time needed to fire the blast is 27.91 s.\

Thus, the blast should be fired at 12.42 s in the -y direction,13.42 s in the +x direction and 27.91 s in the -x direction.

The entire move will require $28.91\mathrm{s}$.