Question

A spaceship maneuvering near planet zeta is located at r⃗ =(600i^−400j^+200k^)×103km, relative to the planet, and traveling at v⃗ =9500i^m/s. it turns on its thruster engine and accelerates with a⃗ =(40i^−20k^)m/s2 for 35 min. part a what is the spaceship's position when the engine shuts off? give your answer as a vector measured in km.

Answer 1

The position of the spaceship as the engine shuts off is given as $\boxed{x=(708.15\,\hat{i}-400\,\hat{j}+155.9\,\hat{k})\times10^3\text{ km}}$.

Explanation:

Given:

The initial position of the spaceship relative to the planet is $(600\,\hat{i}-400\,\hat{j}+200\,\hat{k})\times10^3\text{ km}$.

The speed of the spaceship is $9500\,\hat{i}\text{ m/s}$.

The acceleration of the spaceship is $40\,\hat{i}-20\,\hat{k}\text{ m/s}^2$.

The time for which the thruster engine accelerates the spaceship is $35\text{ min}$.

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.

$\boxed{S=v_{o}t+\dfrac{1}{2}at^2}$

Here, $S$ is the total displacement of the spaceship, $v_{o}$ is the initial velocity of spaceship, $a$ is the acceleration of the spaceship and $t$ 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.

$\begin{aligned}S&=\{9500\,\hat{i}\times(35\times60)\}+\dfrac{1}{2}(40\,\hat{i}-20\,\hat{k})(35\times60)^2\\&=(1.995\times10^7\,\hat{i})+(8.82\times10^7)\,\hat{i}-(4.41\times10^7)\,\hat{k}\\&=(108.15\,\hat{i}-44.1\,\hat{k})\times10^6\text{ m}\\&=(108.15\,\hat{i}-44.1\,\hat{k})\times10^3\text{ km}\end{aligned}$

Write the expression for the new position of the spaceship after it stops.

$X'=X_{o}+S$

Substitute the value of the initial position and the displacement of the spaceship.

$\begin{aligned}X'&=(600\,\hat{i}-400\,\hat{j}+200\,\hat{k})\times10^3\text{ km}+(108.15\,\hat{i}-44.1\,\hat{k})\times10^3\text{ km}\\&=(708.15\,\hat{i}-400\,\hat{j}+155.9\,\hat{k})\times10^3\text{ km}\end{aligned}$

Thus, the position of the spaceship as the engine shuts off is given as $\boxed{x=(708.15\,\hat{i}-400\,\hat{j}+155.9\,\hat{k})\times10^3\text{ km}}$.

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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.

Answer 2

Answer:

 The spaceship's position when the engine shuts off = $(708.15 i - 444.1 j + 200 k)*10^3km$

Explanation:

  Initial location of spaceship = (600 i - 400 j + 200 k)*$10^3km$= (600 i - 400 j + 200 k)*$10^6m$

  Initial velocity = 9500 i m/s

  Acceleration = (40 i - 20 k)$10^3m/s^2$

  Time = 35 minute = 35 * 60 = 2100 seconds

 We have equation of motion , $s= ut+\frac{1}{2} at^2$, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

  Substituting

       $s= (9500 i)*2100+\frac{1}{2}*(40 i - 20 k)*2100^2\\ \\ s=9500*2100 i+20*2100^2i-10*2100^2j\\ \\ s=19.95*10^6i+88.2*10^6i-44.1*10^6j\\ \\ \\s=(108.15i-44.1j)*10^6m$

     So final position = $((600 i - 400 j + 200 k)+(108.15i-44.1j))*10^6=(708.15 i - 444.1 j + 200 k)*10^6m$

                              =$(708.15 i - 444.1 j + 200 k)*10^3km$

    The spaceship's position when the engine shuts off = $(708.15 i - 444.1 j + 200 k)*10^3km$