Question

To simulate artificial gravity, a spaceship accelerates through space at 9.8 m/s2 . a) If it starts from a stop, how long (in time) will it take to reach a speed of 0.1c where c is 3 x 108 m/s (the speed of light). b) How much space does it cover during this time?

Answer 1

The spaceship will take 850.34 hours to reach a speed of 0.1c and it will travel 45 918 367 346.9 km before reaching that speed.

Step-by-step explanation:

The spaceship is traveling at a constant acceleration, then we can use the kinematic equations for that kind of movement to calculate the time it would take to reach a given speed.

`s=s_0+at`

Where
`s_0` is the initial speed,
`a` is the acceleration rate and
`s` is the speed the spaceship will have at a time
`t`.

If the spaceship starts from a stop, then
`s_0=0`. So:

`s=at`

`t=(s)/(a)=(0.1c)/(9.8(m)/(s^2))=(0.1(3*10^8(m)/(s)))/(9.8(m)/(s^2))=3061224.48 s`

`t=3061224.48s*(1 min)/(60 s) *(1 hour)/(60 min) =850.34hours`

Then, the spaceship will take 850.34 hours to reach a speed of 0.1c

To calculate the displacement of the spaceship during that time, we use the following equation:

`x=s_0t+(1)/(2)at^2`

Where
`s_0` is again the initial speed (which is zero),
`a` is the acceleration and
`t` is the travel time (which we've calculated is the previous step).

`x=s_0t+(1)/(2)at^2=(1)/(2)at^2=(1)/(2)(9.8 (m)/(s^2))(3061224.48s)^2=4.59*{10}^(13) m`

`x=4.59*{10}^(13) m * (1 km)/(1000m)=45918367346.9km`

Then, the spaceship will travel 45 918 367 346.9 km before reaching a 0.1c speed.

Answer 2

(a) Time taken to reach the speed will be
`0.306* 10^7sec`

(b) Space cover by spaceship
`0.9183* 10^(14)m`

Step-by-step explanation:

We have given acceleration due to gravity
`g=9.8m/sec^2`

As the spaceship starts from rest so initial velocity u = 0 m/sec

Final velocity v = 0.1 c, here c is speed of light

So final velocity
`v=3* 10^7m/sec`

(a) According to first law of motion we know that v=u+gt, here v is final velocity, u is initial velocity and t is time

So
`3* 10^7=0+9.8* t`

`t=0.306* 10^7sec`

(b) According to third law of motion

`h=ut+(1)/(2)gt^2=0* .306* 10^7+(1)/(2)* 9.8* (0.306* 10^7)^2=0.9183* 10^(14)m`