Suppose a rocket ship in deep space moves with constant acceleration equal to $\mathbf{9}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$, which gives the illusion of normal gravity during the flight. (a)If it starts from rest, how long will it take to acquire a speed one tenth of light (c=$\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$ )? (b)How far will it travel in so doing.

Final velocity of the rocket ship$\frac{1}{10}\mathrm{th}\mathrm{of}\mathrm{the}\mathrm{velocity}\mathrm{of}\mathrm{light}$.

Acceleration of the spaceship (a) $=9.81\mathrm{m}/{\mathrm{s}}^2$.

The problem deals with the kinematic equation of motion in which the motion of an object is described at constant acceleration. The time required to achieve the given speed and distance traveled by the rocket ship can be calculated using kinematic equations.

Formula:

The acceleration in general can be given by,

$a=\frac{v}{t}$ (i)

The displacement in kinematic equation is given by,

$∆x=v_{0}t+\frac{1}{2}a{t}^2$ (ii)

Value of velocity of the rocket ship can be found as follows

$$\begin{gathered} \mathrm{v}=\frac{1}{10}\times 3.0\times {10}^8\frac{\mathrm{m}}{\mathrm{s}} \\ =3.0\times {10}^7\mathrm{m}/\mathrm{s} \end{gathered}$$

To find the time, use the following equation.

$$\begin{gathered} a=\frac{v}{t} \\ t=\frac{v}{a} \\ =\frac{3\times {10}^7}{9.81} \\ =3.1\times {10}^6 \end{gathered}$$

Using equation (ii) the displacement can be written as,

$$\begin{gathered} ∆x=v_{0}t+\frac{1}{2}a{t}^2 \\ =0.5\times a\times t^2 \\ =0.5\times 9.81\times {(3.1\times {10}^6)}^2 \\ =4.6\times {10}^{13}\mathrm{m} \end{gathered}$$