Question

a. How many days will it take a spaceship to accelerate to the speed of light ( 13.0 * 10⁸ m/s) with the acceleration g? b. How far will it travel during this interval? c. What fraction of a light-year is your answer to part b? A light-year is the distance light travels in one year

Short answer

(a) Time required for a spaceship to reach the speed of light is $354days$.

(b) Distance traveled by light in this interval is$4.6\times {10}^{15}m$.

(c) Fraction of light-year in answer to part (b) is $0.49$.

Step-by-step solution

Given information

Acceleration of spaceship is $a=g=9.8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$.

Speed of light$v=c=3\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

Part (a)Time required for a spaceship to reach the speed of light.

Use kinematic equation $v_f=v_i+at$.

If the initial speed of the spaceship is $v_i=0\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

Substitute known values

$$\begin{gathered} 3\times {10}^8=0+9.8\times t \\ t=\frac{3\times {10}^8}{9.8} \\ =30612244.9s \end{gathered}$$

Convert Time in seconds to days.

$$\begin{gathered} t=\frac{30612244.9}{24\times 60\times 60} \\ \approx 354days \end{gathered}$$

Therefore, the time required for a spaceship is$354days$.

Part (b)Distance covered by a spaceship.

Use kinematic equation, $s=v_{i}t+\frac{1}{2}a{t}^2$.

Substitute known values

$$\begin{gathered} s=0+\frac{1}{2}\times 9.8\times {\left(30612244.9\right)}^2 \\ =4.6\times {10}^{15}m \end{gathered}$$

Therefore, the distance covered by a spaceship in 354 days is$4.6\times {10}^{15}m$.

Part (c)

Seconds in one year

$$\begin{gathered} 1yr=365\times 24\times 60\times 60 \\ =31536000s \end{gathered}$$

Distance covered by light in one year is

$$\begin{gathered} s=ct \\ =3\times {10}^8\times 31536000 \\ =9.4608\times {10}^{15}m \end{gathered}$$

Fraction of a light-year in answer to part (b)

$\frac{4.6\times {10}^{15}}{9.4608\times {10}^{15}}=0.49$