A) Well, let's see what happens. The net force acting on the 747 is the braking force, which is constant. We can use Newton's second law (Force = mass x acceleration) to find the acceleration.
Force = mass x acceleration
Rearranging the equation, we get:
Acceleration = Force / mass
Acceleration = (4.30 x 10^5 N) / (3.5 x 10^5 kg)
Calculating this, we find the acceleration is approximately 1.23 m/s^2.
Now, since the force is acting in the opposite direction to the motion of the jetliner, it will decelerate. Using the equation of motion (v = u + at), we can find the velocity (v) of the jetliner 7.50 seconds later.
Given:
Initial velocity, u = 27.0 m/s
Acceleration, a = -1.23 m/s^2 (negative because it's deceleration)
Time, t = 7.50 s
v = u + at
v = 27.0 m/s + (-1.23 m/s^2)(7.50 s)
Calculating this, we find the final velocity (v) of the jetliner after 7.50 seconds is approximately 17.8 m/s.
Therefore, the speed of the jetliner 7.50 seconds later is approximately 17.8 m/s.
B) To find the distance traveled (s) in this time, we can use another equation of motion:
s = ut + (1/2)at^2
Given:
Initial velocity, u = 27.0 m/s
Acceleration, a = -1.23 m/s^2 (negative because it's deceleration)
Time, t = 7.50 s
s = (27.0 m/s)(7.50 s) + (1/2)(-1.23 m/s^2)(7.50 s)^2
Calculating this, we find the distance traveled (s) by the jetliner in 7.50 seconds is approximately 177.2 meters.
Therefore, the jetliner has traveled approximately 177.2 meters in this time.
Hope that adds a bit of amusement to your physics problem!