The space shuttle had a takeoff acceleration approximately 3.5 times the acceleration due to gravity. What was the speed of the space shuttle 120.s after takeoff if it maintained that acceleration?
v = a t = 3.5 * 9.81 * 120 meters/second
so.. is it 4416m/s?
I get 4192
To find the speed of the space shuttle 120 seconds after takeoff, we can use the equation of motion that relates acceleration, time, and velocity:
V = u + at
where:
V is the final velocity
u is the initial velocity (which is 0 in this case, since the shuttle starts from rest)
a is the acceleration
t is the time
Given that the acceleration of the space shuttle during takeoff is approximately 3.5 times the acceleration due to gravity, we can calculate the acceleration:
Acceleration due to gravity (g) = 9.8 m/s^2
Acceleration of the space shuttle = 3.5 * g
Now, we can substitute the values into the equation of motion:
V = 0 + (3.5 * g) * t
V = 3.5 * g * t
To find the speed after 120 seconds, we substitute t = 120 seconds into the equation:
V = 3.5 * 9.8 m/s^2 * 120 s
Now we can calculate the speed using a calculator:
V = 4116 m/s
Therefore, the speed of the space shuttle 120 seconds after takeoff, maintaining an acceleration approximately 3.5 times the acceleration due to gravity, is 4116 m/s.