A small turbo-prop commuter airplane, starting from rest on a Tallahassee airport runway, accelerates for 21.0 s before taking off. Its speed at takeoff is 56.0 m/s (125 mi/hr). Calculate the acceleration of the plane, assuming it remains constant. Express your answer in g's. (i.e., divide the acceleration in m/s2 by 9.81).
To calculate the acceleration of the plane, we can use the formula:
acceleration = change in velocity / time
First, let's convert the speed at takeoff from meters per second (m/s) to kilometers per hour (km/hr) to ensure consistent units:
56.0 m/s * (3.6 km/hr / 1 m/s) = 201.6 km/hr
Next, let's convert the speed from km/hr to meters per second (m/s):
201.6 km/hr * (1000 m / 1 km) * (1 hr / 3600 s) = 56.0 m/s
Now, we can determine the change in velocity by subtracting the initial velocity (0 m/s) from the final velocity (56.0 m/s):
change in velocity = 56.0 m/s - 0 m/s = 56.0 m/s
Finally, we can calculate the acceleration by dividing the change in velocity by the time taken:
acceleration = change in velocity / time = 56.0 m/s / 21.0 s ≈ 2.67 m/s²
To express the acceleration in g's, we divide it by the acceleration due to gravity:
acceleration in g's = acceleration / 9.81 ≈ 2.67 m/s² / 9.81 m/s² ≈ 0.27 g's
Therefore, the acceleration of the plane is approximately 0.27 g's.