Suppose a rocket ship in deep space moves with constant acceleration equal to 9.80 m/s2, which gives the illusion of normal gravity during the flight. (a) If it starts from rest, how long will it take to acquire a speed 12% that of light, which travels at 3.0 × 108 m/s? (b) How far will it travel in so doing
So. I know that 12% is 0.12, I know that acceleration is constant (9.8m/s^2)
I have established that t=0.12
t*a(constant), 0.12*(3.0x10^8)=3.6x10^7.
Apparently my answer is supossed to be 3.6x10^6. I thought the answer should be in seconds, but apparently that is wrong as well.
v= a t
.12 * 3 % 10^8 = 9.8 t
t = .0367 * 10^8 = 3.67 * 10^6
then
d = (1/2) a t^2
To solve this problem, we can use the equations of motion. Let's break it down step by step:
(a) First, we need to find the time it takes for the rocket to reach a speed of 0.12 times the speed of light. We know the acceleration is constant at 9.80 m/s². Let's denote the time as t.
Using the equation of motion: v = u + at, where:
v = final velocity
u = initial velocity (which is 0 m/s since it starts from rest)
a = acceleration
t = time
Substituting the given values into the equation:
0.12(3.0 × 10^8 m/s) = 0 + (9.80 m/s²)t
Simplifying:
3.6 × 10^7 m/s = 9.80 m/s² * t
Now, divide both sides by 9.80 m/s² to isolate t:
t = (3.6 × 10^7 m/s) / (9.80 m/s²)
Calculating this:
t ≈ 3.67 × 10^6 seconds
So, it will take approximately 3.67 million seconds for the rocket to reach a speed 0.12 times that of light.
(b) To find the distance traveled by the rocket during this time, we can use the equation of motion: s = ut + (1/2)at², where:
s = distance
u = initial velocity
t = time
a = acceleration
In this case, the initial velocity (u) is 0 m/s (since the rocket starts from rest).
Substituting the known values into the equation:
s = 0 × t + (1/2) × 9.80 m/s² × (3.67 × 10^6 seconds)²
Simplifying:
s = (1/2) × 9.80 m/s² × (3.67 × 10^6 seconds)²
Calculating this:
s ≈ 6.748 × 10^13 meters
Therefore, the rocket will travel approximately 6.748 × 10^13 meters during this time.
Please note that these calculations assume constant acceleration and neglect any relativistic effects, as the given acceleration allows for normal gravity illusion during the flight.