Suppose a rocket ship in deep space moves with constant acceleration equal to 9.8m/s^2., which gives the illusion of normal gravity duing flight.
a)If it starts from rest, how long will it take to acquire a speed one-tenth that of light, which travels at 3.0x 10^8 m/s?
b) How far will it travel in so doing?
I will be happy to critique your work.
I couldn't figure out which equation I should use.
You should see that
s(t)=(1/2)a*t^2 and v(t)=a*t
a is 9.8m/s^2
Find the time t needed to = .1c with the second equation and use that for t in the first one. Please post work
v(t) = at
.1c= 9.8m/s^2*t
t=.0102s
when I used this t for the first equation I got .000509 butt my book siad the answer was 3.1 x 10^&6 s = 1.2 months
No, c=3.0x 10^8 m/s so .1c=3.0x 10^7 m/s. Thus
3.0x 10^7 m/s=9.8m/s^2*t and t=3.0x 10^7 m/s/9.8m/s^2
You should not get t=.0102s
I got t=3061224.49 and when I pluged t into the second equation I got 4.5918 as the answer
The time looks correct. Keep in mind we only had 2 significant figures to begin with. The distance looks ok too, but I think you're missing an exponent of 10.
It should look like 4.6x10^?
What's the ?(something) here?
would it be 4.6x10^6 ?
Now quit guessing. :)
We know s=(1/2)at^2, v=a*t, a=9.8, c=3x10^8
.1c=3x10^7=9.8*t tells us t is approx. 3x10^6
Thus s=approx. 5x(3x10^6)^2=5*9x10^12=45x10^12 or 4.5x10^13
So it can't possibly be 10^6
To solve this problem, we need to use the equations of motion for uniformly accelerated motion. The first equation relates displacement (s), acceleration (a), and time (t) and is given by:
s = (1/2)at^2
The second equation relates velocity (v), acceleration (a), and time (t) and is given by:
v = at
In this case, the acceleration is equal to 9.8 m/s^2, which is the same as the acceleration due to gravity on Earth. We want to find the time it takes for the rocket ship to acquire a speed equal to one-tenth the speed of light, which is 3.0x10^7 m/s.
a) To find the time (t), we can use the second equation. Rearranging the equation, we have:
t = v/a
Substituting the values, we get:
t = (3.0x10^7 m/s) / (9.8 m/s^2) ≈ 3.06x10^6 s
So, it will take approximately 3.06x10^6 seconds for the rocket ship to acquire a speed one-tenth that of light.
b) To find the distance (s) traveled, we can use the first equation. Substituting the values, we have:
s = (1/2)(9.8 m/s^2)(3.06x10^6 s)^2
Evaluating this expression, we get:
s ≈ 4.5x10^13 m or 4.5x10^6 km
So, the rocket ship will travel approximately 4.5x10^13 meters or 4.5x10^6 kilometers in acquiring a speed one-tenth that of light.
Note: It is important to pay attention to significant figures and units in calculations. In this case, we had to make sure to convert the speed of light from m/s to km/s to match the units used for the acceleration given in the problem.