Respuesta :
Answer:
The spacecraft is now at (1.1 × 10³, 7.2 × 10⁴, 0) km.
Explanation:
Assume that the thruster is the only source of force on the rocket. Acceleration in the first 24.5 seconds:
[tex]\displaystyle a = \frac{\vec{F}}{m}=\frac{<3\times 10^{5},\;0,\;0>}{2.5\times 10^{4}} = <12,\;0,\;0>\;\text{m}\cdot\text{s}^{-2} = <0.012,\;0,\;,0>\;\text{km}\cdot\text{s}^{-2}[/tex].
Distance travelled in the x-direction in the first 24.5 seconds:
[tex]\displaystyle x = \frac{1}{2}\; a\cdot t^{2} = \frac{1}{2} \times 12\times (24.5)^{2} = 3.601\times 10^{3} \;\text{m} = 3.601\;\text{km}[/tex].
Distance travelled in the y-direction in the first 24.5 seconds:
[tex]20\times 24.5 = 490\;\text{km}[/tex].
Velocity at the end of the first 24.5 seconds:
[tex]v = u + a\cdot t = <0,\;20,\;0> + 24.5\times <0.012,\;0,\;0> =<0.294,\;20,\;0>\;\text{km}\cdot\text{s}^{-1}[/tex].
Velocity of the spacecraft is constant between 24.5 seconds and 3600 seconds since there's no external force on the spacecraft. Displacement in that period of time:
[tex]x = v\cdot t = (3600-24.5)\cdot <0.294,\;20,\;0> = <1051,\;71510,\;0>\;\text{km}[/tex].
Position vector of the spacecraft:
- Initial: <6, 7, 0> km;
- After 24.5 seconds: <6, 7, 0> + <3.601, 490, 0> = <9.601, 497, 0> km;
- After 3600 seconds: <9.601, 497, 0> + <1051, 71510, 0> = <1060, 72007, 0> km.
Round to an appropriate number of significant figures. Position of the spacecraft after 3600 seconds: (1.1 ×10³, 7.2 × 10⁴, 0) km.
The final position of the spacecraft is [tex]\vec r = <1064.820, 72497, 0>\,[km][/tex].
Let be a particle whose initial conditions are represented by vectors [tex]\vec r_{o}[/tex], in meters and [tex]\vec v_{o}[/tex], in meters per second. From statement we understand that acceleration ([tex]\vec a[/tex]), in meters per square second, is of the form:
[tex]\vec a = \frac{1}{m}\cdot \vec F[/tex] (1)
Where:
[tex]\vec F[/tex] - Vector force, in newtons.
[tex]m[/tex] - Mass, in kilograms.
Please notice that force is uniform in time.
By definitions of Position, Velocity and Differential Calculus, we have the following parametric equations for the velocity and position of the spacecraft in time:
[tex]\vec v = \vec v_{o} + \frac{t}{m}\cdot \vec F[/tex] (2)
[tex]\vec r = \vec r_{o} + t\cdot \vec v_{o} +\frac{t^{2}}{2\cdot m}\cdot \vec F[/tex] (3)
Where [tex]t[/tex] is the time, in seconds.
There are two stages in the motion of spacecraft:
1) Thruster rockets are turned on for 24.5 seconds. ([tex]\vec F \ne \vec O[/tex]).
2) Thruster rockets are turned off. ([tex]\vec F = \vec O[/tex]).
Stage 1:
([tex]\vec r_{o} = <6000, 7000, 0>\,\left[\frac{m}{s} \right][/tex], [tex]\vec v_{o} = <0, 20000, 0>\,\left[\frac{m}{s} \right][/tex], [tex]\vec F = <3\times 10^{5}, 0, 0>\,[N][/tex], [tex]t = 24.5\,s[/tex], [tex]m = 2.5\times 10^{4}\,kg[/tex])
By (2):
[tex]\vec v = <0, 20000, 0> + \left(\frac{24.5\,s}{2.5\times 10^{4}\,kg}\right)\cdot <3\times 10^{5}, 0, 0>[/tex]
[tex]\vec v = <294, 20000, 0>\,\left[\frac{m}{s}\right][/tex]
By (1):
[tex]\vec r = <6000, 7000, 0> + (24.5)\cdot <0, 20000, 0> + \left[\frac{(24.5)^{2}}{2\cdot (2.5\times 10^{4})} \right]\cdot <3\times 10^{5}, 0, 0>[/tex]
[tex]\vec r = <6420.175, 497000, 0>\,\left[m\right][/tex]
Stage 2:
([tex]\vec r_{o} = <6420.175, 497000, 0>\,\left[m\right][/tex], [tex]\vec v_{o} = <294, 20000, 0>\,\left[\frac{m}{s}\right][/tex], [tex]\vec F = <0,0,0>\,\left[N \right][/tex], [tex]t = 3600\,s[/tex], [tex]m = 2.5\times 10^{4}\,kg[/tex])
By (2):
[tex]\vec v = <294, 20000, 0>\,\left[\frac{m}{s}\right][/tex]
By (1):
[tex]\vec r = <6420.175, 497000, 0> + (3600)\cdot <294, 20000,0>[/tex]
[tex]\vec r = <1064820.175, 72497000, 0>\,[m][/tex]
[tex]\vec r = <1064.820, 72497, 0>\,[km][/tex]
The final position of the spacecraft is [tex]\vec r = <1064.820, 72497, 0>\,[km][/tex].
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