an airplane starts from rest and accelerates at a constant 3.00m/s2 for 30.0s before leaving the ground.
a). how far did it move?
b). how fast was the airplane going when it took off?
An airport runaway is 150 n long an airplane accelerates at 2 m/ sec 2
a, what is the final speed of the plane if it covers the entire runaway?
Why did the airplane go to school? To get a higher-education in takeoff physics! Now, let's calculate the answers to your questions.
a) To determine how far the airplane moved, we can use the equation: distance = (0.5) * acceleration * time^2. Plugging in the values:
distance = (0.5) * 3.00 m/s^2 * (30.0 s)^2, we find that the airplane moved 1350.0 meters.
b) Fun fact - airplanes hate telling you their speed since it's a sensitive topic for them. But luckily, we can calculate it! We can use the equation: final velocity = initial velocity + acceleration * time. Since the airplane starts from rest, the initial velocity is 0 m/s. Plugging in the values, we have:
final velocity = 0 m/s + 3.00 m/s^2 * 30.0 s = 90.0 m/s.
So, the airplane was flying at a speed of 90.0 m/s when it took off. Quite an impressive takeoff speed, I must say!
To answer these questions, we need to use the kinematic equations of motion. The key information given in the problem is the initial velocity (0 m/s), acceleration (3.00 m/s^2), and time (30.0 s).
a). To find the distance moved by the airplane, we can use the following kinematic equation:
d = vit + 0.5at^2
where
d = distance moved
vi = initial velocity (0 m/s)
a = acceleration (3.00 m/s^2)
t = time (30.0 s)
Substituting the given values into the equation:
d = (0 m/s)(30.0 s) + 0.5(3.00 m/s^2)(30.0 s)^2
Simplifying:
d = 0 + 0.5(3.00 m/s^2)(900.0 s^2)
d = 0 + 1350.0 m^2/s^2
d = 1350.0 m
Therefore, the airplane moved a distance of 1350.0 meters.
b). To find the final velocity when the airplane took off, we can use the following kinematic equation:
vf = vi + at
where
vf = final velocity (unknown)
vi = initial velocity (0 m/s)
a = acceleration (3.00 m/s^2)
t = time (30.0 s)
Substituting the given values into the equation:
vf = (0 m/s) + (3.00 m/s^2)(30.0 s)
Simplifying:
vf = 0 m/s + 90.0 m/s
vf = 90.0 m/s
Therefore, the airplane was going at a speed of 90.0 meters per second when it took off.
An airplane starts from rest and accelerates east at a constant rate of 3 m/s2 for 30 seconds.
What is the displacement
a. d = 0.5a*t^2.
d = 0.5*3*30^2 = 1350 m.
b. V = a*t = 3 * 30 = 90 m/s.