Respuesta :
Answer:
The spaceship's position when the engine shuts off = [tex](708.15 i - 444.1 j + 200 k)*10^3km[/tex]
Explanation:
Initial location of spaceship = (600 i - 400 j + 200 k)*[tex]10^3km[/tex]= (600 i - 400 j + 200 k)*[tex]10^6m[/tex]
Initial velocity = 9500 i m/s
Acceleration = (40 i - 20 k)[tex]10^3m/s^2[/tex]
Time = 35 minute = 35 * 60 = 2100 seconds
We have equation of motion , [tex]s= ut+\frac{1}{2} at^2[/tex], s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
Substituting
[tex]s= (9500 i)*2100+\frac{1}{2}*(40 i - 20 k)*2100^2\\ \\ s=9500*2100 i+20*2100^2i-10*2100^2j\\ \\ s=19.95*10^6i+88.2*10^6i-44.1*10^6j\\ \\ \\s=(108.15i-44.1j)*10^6m[/tex]
So final position = [tex]((600 i - 400 j + 200 k)+(108.15i-44.1j))*10^6=(708.15 i - 444.1 j + 200 k)*10^6m[/tex]
=[tex](708.15 i - 444.1 j + 200 k)*10^3km[/tex]
The spaceship's position when the engine shuts off = [tex](708.15 i - 444.1 j + 200 k)*10^3km[/tex]
The position of the spaceship as the engine shuts off is given as [tex]\boxed{x=(708.15\,\hat{i}-400\,\hat{j}+155.9\,\hat{k})\times10^3\text{ km}}[/tex].
Explanation:
Given:
The initial position of the spaceship relative to the planet is [tex](600\,\hat{i}-400\,\hat{j}+200\,\hat{k})\times10^3\text{ km}[/tex].
The speed of the spaceship is [tex]9500\,\hat{i}\text{ m/s}[/tex].
The acceleration of the spaceship is [tex]40\,\hat{i}-20\,\hat{k}\text{ m/s}^2[/tex].
The time for which the thruster engine accelerates the spaceship is [tex]35\text{ min}[/tex].
Concept:
As the spaceship is accelerated by the thruster engine, the spaceship moves according to the second equation of motion.
Write the expression for the displacement of the spaceship in time interval.
[tex]\boxed{S=v_{o}t+\dfrac{1}{2}at^2}[/tex]
Here, [tex]S[/tex] is the total displacement of the spaceship, [tex]v_{o}[/tex] is the initial velocity of spaceship, [tex]a[/tex] is the acceleration of the spaceship and [tex]t[/tex] is the time for which the spaceship accelerates.
Substitute the values of the initial speed, acceleration and the time taken by the spaceship in above expression.
[tex]\begin{aligned}S&=\{9500\,\hat{i}\times(35\times60)\}+\dfrac{1}{2}(40\,\hat{i}-20\,\hat{k})(35\times60)^2\\&=(1.995\times10^7\,\hat{i})+(8.82\times10^7)\,\hat{i}-(4.41\times10^7)\,\hat{k}\\&=(108.15\,\hat{i}-44.1\,\hat{k})\times10^6\text{ m}\\&=(108.15\,\hat{i}-44.1\,\hat{k})\times10^3\text{ km}\end{aligned}[/tex]
Write the expression for the new position of the spaceship after it stops.
[tex]X'=X_{o}+S[/tex]
Substitute the value of the initial position and the displacement of the spaceship.
[tex]\begin{aligned}X'&=(600\,\hat{i}-400\,\hat{j}+200\,\hat{k})\times10^3\text{ km}+(108.15\,\hat{i}-44.1\,\hat{k})\times10^3\text{ km}\\&=(708.15\,\hat{i}-400\,\hat{j}+155.9\,\hat{k})\times10^3\text{ km}\end{aligned}[/tex]
Thus, the position of the spaceship as the engine shuts off is given as [tex]\boxed{x=(708.15\,\hat{i}-400\,\hat{j}+155.9\,\hat{k})\times10^3\text{ km}}[/tex].
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Answer Details:
Grade: High School
Subject: Physics
Chapter: Scalar and Vector
Keywords:
spaceship, maneuvering, planet zeta, travelling, thruster, displacement, acceleration, initial velocity, engine, time, position.
