Starting from rest an airplane moves with a constant acceleration of 2m/s for 30 seconds before it takes off
A. How far does it travel down the runway before it takes off ?
B. What's it's final velocity ?
A. d = Vo*t + 0.5a*t^2.
d = 0 + 1*30^2 =
B. V = a*t.
1.00m
To find the answers to these questions, we can use the equations of motion.
A. The distance traveled, also known as displacement, can be found using the equation:
displacement = initial velocity * time + (1/2) * acceleration * time^2
In this case, the initial velocity is zero since the airplane starts from rest. The acceleration is given as 2 m/s^2, and the time is 30 seconds. Plugging these values into the equation:
displacement = 0 * 30 + (1/2) * 2 * (30^2)
= 0 + 1 * (900)
= 900 meters
Therefore, the airplane travels 900 meters down the runway before it takes off.
B. The final velocity can be found using the equation:
final velocity = initial velocity + acceleration * time
Again, the initial velocity is zero. Plugging in the given values:
final velocity = 0 + 2 * 30
= 60 m/s
Therefore, the final velocity of the airplane is 60 m/s.