A $8.0\times {10}^{4}kg$spaceship is at rest in deep space. Its thrusters provide a force of $1200kN.$The spaceship fires its thrusters for $20$s, then coasts for$12$ km. How long does it take the spaceship to coast this distance?

The force exerted on a mass that causes it to change velocity.

Newton's first law of motion states that an item remains at rest or in uniform motion until it is acted with by a net external force.

The quantity of force exerted on an object is equal to the product of its mass and the acceleration generated, according to Newton's second equation of motion.

$F=ma$

The final velocity$\left(v\right)$of an object with starting velocity$\left(u\right)$and accelerated with acceleration$\left(a\right)$for time$\left(t\right)$is described by kinematics first law as $v=u+at$.

The object in this scenario is a spacecraft on which a force is applied first, causing its velocity to grow, and then the force is removed, and the spaceship moves freely in a uniform motion.

Spaceship mass $\left(m\right)=8.0\times {10}^{4}kg$,

Thrusters force $\left(F\right)=1200\times {10}^{3}N$,

Force time applied $\left(t_1\right)=20s$,

Coasting distance $\left(d\right)=12km=12\times {10}^{3}m$,

Initial velocity $\left(u\right)=0m/s$.

Formula for acceleration, $a=\frac{F}{m}$

$$\begin{gathered} a=\frac{1200\times {10}^3}{8.0\times {10}^4} \\ a=15m/s^2 \end{gathered}$$

Formula for final velocity, $v=u+a{t}_1$

$$\begin{gathered} v=0+15\times 20 \\ v=300m/s \end{gathered}$$

Formula for coasting time, $t=\frac{d}{v}$

$$\begin{gathered} t=\frac{12\times {10}^3}{300} \\ t=40s \end{gathered}$$

Hence, moving with uniform velocity it takes$t=40s$to coast$12km$.