An airplane requires 420 m of runway to take off from an airstrip. If the plane starts from rest and it takes 32 s to lift off, with what velocity does it leave the ground? What was the average acceleration of the plane?
d = .5 a t^2
840 = a(1024)
a = .820 m/s^2
v = a t = .820 * 32 = 26.2 m/s
To find the velocity with which the airplane leaves the ground, we can use the formula:
velocity = distance / time.
In this case, the distance is the length of the runway required for takeoff (420 m), and the time is the time taken to lift off (32 s). So, the velocity can be calculated as:
velocity = 420 m / 32 s
Simplifying this equation gives:
velocity = 13.125 m/s
Therefore, the airplane leaves the ground with a velocity of 13.125 m/s.
To find the average acceleration of the plane, we can use the formula:
acceleration = change in velocity / time taken.
Since the plane starts from rest, the initial velocity is 0 m/s. So, the change in velocity is the final velocity (calculated above) minus the initial velocity:
change in velocity = final velocity - initial velocity = 13.125 m/s - 0 m/s = 13.125 m/s.
The time taken for takeoff is given as 32 s.
Therefore, the average acceleration can be calculated as:
acceleration = 13.125 m/s / 32 s
Simplifying this equation gives:
acceleration = 0.410 m/s²
Therefore, the average acceleration of the plane is 0.410 m/s².