Question

Suppose a rocket ship in deep space moves with a constant acceleration equal to $9.8m{{s}^{-2}}$ , which gives the illusion of normal gravity during the flight. If it starts from rest, how long will it take to acquire a speed one-tenth that of light, which travels at $3.0\times {{10}^{8}}m{{s}^{-1}}$? How far will it travel in doing so?

Answer 1

Hint: Here we need to appropriately apply the equations of the kinematics. The initial velocity, final velocity and the acceleration of the rocket is given. The rocket is moving with uniform acceleration, so we can apply the kinematics equations here. First we will determine the required time, then we will calculate the distance travelled by the rocket.

Formula used:
 $\begin{align}
  & v=u+at,s=ut+\dfrac{1}{2}a{{t}^{2}} \\
 & \\
\end{align}$

Complete step by step answer:
Here the initial velocity $u$ of the rocket is $0$ . the final velocity $v$ is given by
$v=\dfrac{3\times {{10}^{8}}}{10}=3\times {{10}^{7}}m{{s}^{-1}}$ and the acceleration$a=9.8m{{s}^{-2}}$ . Now if $t$ is the required time, then from the kinematics equation we can write
$\begin{align}
  & v=u+at \\
 & or3\times {{10}^{7}}=0+9.8\times t \\
 & ort=\dfrac{3\times {{10}^{7}}}{9.8}=3.06\times {{10}^{6}}s \\
\end{align}$
So it will take $3.06\times {{10}^{6}}s.$
Now if $s$ is the distance travelled in time $t$ then from the kinematics equation we can write
$\begin{align}
  & s=ut+\dfrac{1}{2}a{{t}^{2}} \\
 & ors=0\times 3.06\times {{10}^{6}}+\dfrac{1}{2}\times 9.8\times {{(3.06\times {{10}^{6}})}^{2}} \\
 & ors=4.59\times {{10}^{13}}m \\
\end{align}$

So it will travel a distance of $4.59\times {{10}^{13}}m$.

Note: To calculate the distance travelled we could have used the relation ${{v}^{2}}={{u}^{2}}+2as$ . As all the other quantities are known to us we can get the value of $s.$ Here we are able to apply the kinematics equations because the rocket is travelling with uniform acceleration. If the rocket was travelling with non uniform acceleration, the kinematics equations could not have been applied . Here we also have to remember that all the quantities should be converted in the same system of units. Here all the quantities are given in the SI system so we will have the value of time and distance in seconds and in meters respectively.