A 1200 kg airplane starts from rest and with a constant average forward acceleration of magbitude 5.0 m/s, reaches its takeoff speed in 9.00 s. The runway is 300 m long. What is the plane's takeoff speed?
Why did the airplane start taking acting lessons?
Because it wanted to learn how to 'take off'!
To find the plane's takeoff speed, we can use the kinematic equation:
v = u + at
Where:
v = final velocity (takeoff speed)
u = initial velocity (0 m/s since the plane starts from rest)
a = acceleration (5.0 m/s²)
t = time (9.00 s)
First, we need to determine the distance traveled by the plane during the acceleration phase. We can use the equation:
s = ut + 0.5at²
Where:
s = distance
u = initial velocity
t = time
a = acceleration
Given that the initial velocity is 0 m/s and the time is 9.00 s, we can calculate the distance using the equation:
s = 0 + 0.5 * 5.0 * (9.00^2)
s = 0.5 * 5.0 * 81.0
s = 202.5 m
Now, we can calculate the final velocity (takeoff speed) using the equation:
v = u + at
v = 0 + 5.0 * 9.00
v = 45.0 m/s
Therefore, the plane's takeoff speed is 45.0 m/s.
To find the plane's takeoff speed, we can use the equations of motion. The first equation we will use is:
v = u + at
where,
v = final velocity (takeoff speed),
u = initial velocity (0 m/s),
a = acceleration,
t = time taken (9.00 s).
Substituting the given values into the equation:
v = 0 + (5.0 m/s^2) * (9.00 s)
v = 45 m/s
So, the plane's takeoff speed is 45 m/s.
The acceleration rate times the time spent taking off is the takeoff speed. Two of the numbers you were given (mass and runway length) are not needed to solve the problem. Takeoff does not require the full runway length. It would be dangerous if it did.
The runway length actually used is
(1/2) a t^2 = (1/2)*5*9^2 = 202.5 m