Solveeit Logo

Question

Question: A spacecraft flying in a straight course with a velocity of \[60\,km{s^{ - 1}}\] fires its rocket mo...

A spacecraft flying in a straight course with a velocity of 60kms160\,km{s^{ - 1}} fires its rocket motors for 5s5\,s. At the end of this time, its speed is 90kms190\,km{s^{ - 1}} in the same direction. Find the distance traveled by spacecraft in the first 8s8\,s after the rocket motors were started, motors have been in action for only 5s5\,s.

Explanation

Solution

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+atv = u + at
v2=u2+2as\Rightarrow {v^2} = {u^2} + 2as
Where, u=u = Initial velocity, v=v = final velocity, t=t = time, a=a = acceleration, and s=s = distance.

Complete step by step answer:
From the given question, we know that the initial velocity is 60 kms1km{s^{ - 1}}and the final velocity is 90kms190km{s^{ - 1}}. Assuming that after 5s5s, the rocket stops and is in space, and due to inertia it continues to move at a constant velocity of 90 kms1km{s^{ - 1}}. Thus, after 5s5s the rocket travels 33 more seconds.

Given data,
u=60kms1u = 60km{s^{ - 1}}
v=90kms1\Rightarrow v = 90km{s^{ - 1}}
t=5s\Rightarrow t = 5s
Using these data,
v=u+atv = u + at
90=60+5a\Rightarrow 90 = 60 + 5a...........[acceleration = aa ]
5a=9060\Rightarrow 5a = 90 - 60
a=305=6kms2\Rightarrow a = \dfrac{{30}}{5} = 6\,km{s^{ - 2}}

Now, from the formula v2=u2+2as{v^2} = {u^2} + 2as , the distance can be found with the uniform acceleration.
v2=u2+2as{v^2} = {u^2} + 2as
s=v2u22a\Rightarrow s = \dfrac{{{v^2} - {u^2}}}{{2a}}
s=9026022×6\Rightarrow s = \dfrac{{{{90}^2} - {{60}^2}}}{{2 \times 6}}
s=(90+60)(9060)12\Rightarrow s = \dfrac{{\left( {90 + 60} \right)\left( {90 - 60} \right)}}{{12}}
s=150×3012\Rightarrow s = \dfrac{{150 \times 30}}{{12}}
s=375km\Rightarrow s = 375\,km

Now, since the craft moves farther due to its inertia, it can be assumed that it moves with uniform velocity 90kms190km{s^{ - 1}} for another 3s3s,
Hence, the distance will be covered by the uniform velocity is,
s=vts = vt
s=120×3\Rightarrow s = 120 \times 3
s=360km\therefore s = 360km

So, the total distance covered by the spacecraft in 8s8s is (375+360)km=735km(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.