Physics, 09.01.2020 20:31 Nico8776

Suppose you are navigating a spacecraft far from other objects. the mass of the spacecraft is 2.5 × 104 kg (about 25 tons). the rocket engines are shut off, and you're coasting along with a constant velocity of < 0, 20, 0> km/s. as you pass the location < 6, 7, 0> km you briefly fire side thruster rockets, so that your spacecraft experiences a net force of < 3 × 105, 0, 0> n for 24.5 s. the ejected gases have a mass that is small compared to the mass of the spacecraft. you then continue coasting with the rocket engines turned off. where are you an hour later? (think about what approximations or simplifying assumptions you made in your analysis. also think about the choice of system: what are the surroundings that exert external forces on your system? )

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Answer from: meaghan18

The final position of the spacecraft is $\vec r = \,[km]$ .

Let be a particle whose initial conditions are represented by vectors $\vec r_{o}$ , in meters and $\vec v_{o}$ , in meters per second. From statement we understand that acceleration ( $\vec a$ ), in meters per square second, is of the form:

$\vec a = \frac{1}{m}\cdot \vec F$ (1)

Where:

$\vec F$ - Vector force, in newtons.

- 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:

$\vec v = \vec v_{o} + \frac{t}{m}\cdot \vec F$ (2)

$\vec r = \vec r_{o} + t\cdot \vec v_{o} +\frac{t^{2}}{2\cdot m}\cdot \vec F$ (3)

Where is the time, in seconds.

There are two stages in the motion of spacecraft:

1) Thruster rockets are turned on for 24.5 seconds. ( $\vec F \ne \vec O$ ).

2) Thruster rockets are turned off. ( $\vec F = \vec O$ ).

Stage 1:

( $\vec r_{o} = \,\left[\frac{m}{s} \right]$ , $\vec v_{o} = \,\left[\frac{m}{s} \right]$ , $\vec F = \,[N]$ , $t = 24.5\,s$ , $m = 2.5\times 10^{4}\,kg$ )

By (2):

$\vec v = + \left(\frac{24.5\,s}{2.5\times 10^{4}\,kg}\right)\cdot$

$\vec v = \,\left[\frac{m}{s}\right]$

By (1):

$\vec r = + (24.5)\cdot + \left[\frac{(24.5)^{2}}{2\cdot (2.5\times 10^{4})} \right]\cdot$

$\vec r = \,\left[m\right]$

Stage 2:

( $\vec r_{o} = \,\left[m\right]$ , $\vec v_{o} = \,\left[\frac{m}{s}\right]$ , $\vec F = \,\left[N \right]$ , $t = 3600\,s$ , $m = 2.5\times 10^{4}\,kg$ )

By (2):

$\vec v = \,\left[\frac{m}{s}\right]$

By (1):

$\vec r = + (3600)\cdot$

$\vec r = \,[m]$

$\vec r = \,[km]$

The final position of the spacecraft is $\vec r = \,[km]$ .

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Answer from: mahmudabiazp3ekot

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:

$\displaystyle a = \frac{\vec{F}}{m}=\frac{}{2.5\times 10^{4}} = \;\text{m}\cdot\text{s}^{-2} = \;\text{km}\cdot\text{s}^{-2}$ .

Distance travelled in the x-direction in the first 24.5 seconds:

$\displaystyle x = \frac{1}{2}\; a\cdot t^{2} = \frac{1}{2} \times 12\times (24.5)^{2} = 3.601\times 10^{3} \;\text{m} = 3.601\;\text{km}$ .

Distance travelled in the y-direction in the first 24.5 seconds:

$20\times 24.5 = 490\;\text{km}$ .

Velocity at the end of the first 24.5 seconds:

$v = u + a\cdot t = + 24.5\times =\;\text{km}\cdot\text{s}^{-1}$ .

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:

$x = v\cdot t = (3600-24.5)\cdot = \;\text{km}$ .

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.

Answer from: Quest

the answer is b, have negligible mass

Answer from: Quest

answer: a. same unit

explanation:

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