Question

A spaceship moving with an initial velocity of 58.0 meters/second experiences a uniform acceleration and attains a final velocity of 153 meters/second. What distance has the spaceship covered after 12.0 seconds?

6.96 × 102 meters
1.27 × 103 meters
5.70 × 102 meters
1.26 × 102 meters
6.28 × 102 meters

Answer 1

Final distance is 1266 meters

Answer 2

Distance,
`s=1.27* 10^3\ meters`

Step-by-step 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 :

`a=(v-u)/(t)`

`a=(153-58)/(12)`

`a=7.91\ m/s^2`

Now use third equation of motion as :

`s=(v^2-u^2)/(2a)`

`s=((153)^2-(58)^2)/(2* 7.91)`

s = 1267.06 meters

or

`s=1.27* 10^3\ meters`

So, the distance covered by the spaceship is
`1.27* 10^3\ meters`. Hence, this is the required solution.