Answer:
Distance, [tex]s=1.27\times 10^3\ meters[/tex]
Explanation:
Given that,
Initial velocity of the spaceship, u = 58 m/s
Final speed of the spaceship, v = 153 m/s
Time, t = 12 s
We need to find the distance covered by the spaceship after 12 seconds. Let it is equal to s. Using first equation of motion as :
[tex]a=\dfrac{v-u}{t}[/tex]
[tex]a=\dfrac{153-58}{12}[/tex]
[tex]a=7.91\ m/s^2[/tex]
Now use third equation of motion as :
[tex]s=\dfrac{v^2-u^2}{2a}[/tex]
[tex]s=\dfrac{(153)^2-(58)^2}{2\times 7.91}[/tex]
s = 1267.06 meters
or
[tex]s=1.27\times 10^3\ meters[/tex]
So, the distance covered by the spaceship is [tex]1.27\times 10^3\ meters[/tex]. Hence, this is the required solution.
