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? )

Answers: 2

Show answers

Answers

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 $\vec r = \,[km]$ .

Link to a related question: link

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

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s

Answer from: Quest

An atom that has fewer electrons that protons is called a positive ion.

Another question on Physics

Physics, 22.06.2019 05:00

Aperson walking 1 mile everyday for exercise, leaving her front porch at 9: 00 am and returning to her front porch at 9: 25 am. what is the total displacement of her daily walk? a) 1 mileb) 0c) 25 minutes d) none of the above

Answers: 1

Answer

Physics, 22.06.2019 15:30

Two point charges are separated by a certain distance. how does the strength of the electric field produced by the first charge, at the position of the second charge, change if the second charge is doubled? a. the field does not change b. the field strength decreases by half. c. the field strength doubles d. the field strength quadruples

Answers: 1

Answer

Physics, 22.06.2019 17:20

Debris produced by mining is called

Answers: 1

Answer

Physics, 22.06.2019 18:30

Ben (55kg) is standing on very slippery ice when junior (25kg) bumps into him. junior was moving at a speed of 8 m/s before the collision and ben and junior embrace after the collision. find the speed of ben and junior as they move across the ice after the collision. give the answer in m/s. describe the work you did to get the answer.

Answers: 1

Answer

You know the right answer?

Suppose you are navigating a spacecraft far from other objects. the mass of the spacecraft is 2.5 ×...

Questions