Question

A spacecraft flying in a straight course with a velocity of \[60\,km{s^{ - 1}}\] fires its rocket motors for \[5\,s\]. At the end of this time, its speed is \[90\,km{s^{ - 1}}\] in the same direction. Find the distance traveled by spacecraft in the first \[8\,s\] after the rocket motors were started, motors have been in action for only \[5\,s\].

Answer 1

Hint: Find the acceleration of a spacecraft flying in a straight course by using the law of motion formula. Use it to find the distance traveled by spacecraft in the first eight seconds after the rocket motors were started, motors have been in action for only five seconds.

Formula used:
\[v = u + at\]
\[\Rightarrow {v^2} = {u^2} + 2as\]
Where, \[u = \]Initial velocity, \[v = \]final velocity, \[t = \]time, \[a = \]acceleration, and \[s = \]distance.

Complete step by step answer:
From the given question, we know that the initial velocity is 60 \[km{s^{ - 1}}\]and the final velocity is \[90km{s^{ - 1}}\]. Assuming that after \[5s\], the rocket stops and is in space, and due to inertia it continues to move at a constant velocity of 90 \[km{s^{ - 1}}\]. Thus, after \[5s\] the rocket travels \[3\] more seconds.

Given data,
\[u = 60km{s^{ - 1}}\]
\[\Rightarrow v = 90km{s^{ - 1}}\]
\[\Rightarrow t = 5s\]
Using these data,
\[v = u + at\]
\[ \Rightarrow 90 = 60 + 5a\]...........[acceleration = $a$ ]
\[ \Rightarrow 5a = 90 - 60\]
\[ \Rightarrow a = \dfrac{{30}}{5} = 6\,km{s^{ - 2}}\]

Now, from the formula \[{v^2} = {u^2} + 2as\] , the distance can be found with the uniform acceleration.
\[{v^2} = {u^2} + 2as\]
$ \Rightarrow s = \dfrac{{{v^2} - {u^2}}}{{2a}}$
$ \Rightarrow s = \dfrac{{{{90}^2} - {{60}^2}}}{{2 \times 6}}$
$ \Rightarrow s = \dfrac{{\left( {90 + 60} \right)\left( {90 - 60} \right)}}{{12}}$
$ \Rightarrow s = \dfrac{{150 \times 30}}{{12}}$
$\Rightarrow s = 375\,km$

Now, since the craft moves farther due to its inertia, it can be assumed that it moves with uniform velocity \[90km{s^{ - 1}}\] for another \[3s\],
Hence, the distance will be covered by the uniform velocity is,
$s = vt$
$ \Rightarrow s = 120 \times 3$
$\therefore s = 360km$

So, the total distance covered by the spacecraft in $8s$ is $(375 + 360)km = 735 km$.

Note: Spacecrafts need to be equipped with an array of features so that the crew inside them can be safe and work properly. The distance and duration undertaken by a spacecraft demand a reliable system that can be operated far from home. The system should be lightweight so that the rocket can carry it. Most importantly, it should provide all the emergency needs of the astronauts.