An airplane starts from rest and accelerates at a constant rate if 3.00m/s^2 for 30.0s before leaving the ground.
A)How far did it move?
B) how fast was it going when it took off?
d=1/2 a t^2
vf=a*t
a. 1350m
To find the answers to these questions, we can use the equations of motion.
A) To find how far the airplane moved, we can use the equation:
distance = initial velocity * time + (1/2) * acceleration * time^2
Since the airplane started from rest, the initial velocity is 0. The time is given as 30.0s, and the acceleration is given as 3.00m/s^2. Plugging these values into the equation, we can calculate the distance:
distance = 0 * 30.0 + (1/2) * 3.00 * (30.0)^2
Simplifying the equation:
distance = 0 + (1/2) * 3.00 * 900
distance = 0 + 450
Therefore, the airplane moved a distance of 450 meters.
B) To find the speed of the airplane when it took off, we can use the equation:
final velocity = initial velocity + acceleration * time
The initial velocity is 0, the acceleration is 3.00m/s^2, and the time is given as 30.0s. Plugging these values into the equation, we can calculate the final velocity:
final velocity = 0 + 3.00 * 30.0
final velocity = 0 + 90.0
Therefore, the airplane was going at a speed of 90.0 m/s when it took off.