A 747 jetliner lands and begins to slow to a stop as it moves along the runway. The mass is 3.1 105 kg, its speed is 25.5 m/s, and the net braking force is 4.24 105 N.(a) What is its speed 7.45 s later?How far has it traveled in this time?
a. ma,
a = F/m = -.24*10^5 / 3.1*10^5 = -.37m/s^2.
Vf = Vo + at,
Vf = 25.5 -1.37*7.45 = 15.29m/s.
b. d = (Vf^2 - Vo^2) / 2a,
d = ((15.29)^2 - (25.5)^2) / -2.74 =
152m.
To find the speed of the jetliner 7.45 seconds later, we can use the equation for acceleration:
acceleration = (force / mass)
Given that the net braking force is 4.24 * 10^5 N and the mass is 3.1 * 10^5 kg, we can substitute these values into the equation:
acceleration = (4.24 * 10^5 N) / (3.1 * 10^5 kg)
Now we can calculate the acceleration:
acceleration = 1.37 m/s^2
To find the final speed of the jetliner after 7.45 seconds, we can use the equation:
final speed = initial speed + (acceleration * time)
Given that the initial speed is 25.5 m/s and the time is 7.45 seconds, we can substitute these values into the equation:
final speed = 25.5 m/s + (1.37 m/s^2 * 7.45 s)
final speed = 35.55 m/s
Therefore, the speed of the jetliner 7.45 seconds later is 35.55 m/s.
To find the distance traveled by the jetliner in this time, we can use the equation:
distance = (initial speed * time) + (0.5 * acceleration * time^2)
Given that the initial speed is 25.5 m/s, the time is 7.45 seconds, and the acceleration is 1.37 m/s^2, we can substitute these values into the equation:
distance = (25.5 m/s * 7.45 s) + (0.5 * 1.37 m/s^2 * (7.45 s)^2)
distance = 189.98 m + 57.89 m
distance = 247.87 m
Therefore, the jetliner has traveled a distance of 247.87 meters in this time.