To take off from the ground, an airplane must reach a sufficiently high speed. The velocity required for the takeoff, the takeoff velocity, depends on several factors, including the weight of the aircraft and the wind velocity.
Part A: A plane accelerates from rest at a constant rate of 5.00m/s2 along a runway that is 1800m long. Assume that the plane reaches the required takeoff velocity at the end of the runway. What is the time tTO needed to take off?
(the answer is 26.8s; i need help only on part B)
Part B: What is the speed vTO of the plane as it takes off?
time*acceleration=134m/s
134 m/s a=vt a=26.83s•5.00m/s^2
a)s = ut + 0.5at^2
s = distance = 1800m, u = initial velocity = 0, a = acceleration = 5m/s^2, t = time
1800 = 0 + 0.5*5*t^2
t^2 = 1800/2.5 = 720
t = 26.83 s (to 4 sig figs)
To find the speed vTO of the plane as it takes off, we can use the kinematic equation:
v = u + at
where:
v is the final velocity (takeoff velocity),
u is the initial velocity (which is zero since the plane starts from rest),
a is the acceleration (5.00 m/s^2 in this case),
and t is the time taken to reach the final velocity.
Since we already know the acceleration and we're looking for the final velocity, we need to find the time taken to reach that velocity.
In Part A, we found that the plane takes 26.8 seconds (tTO) to take off. Thus, we can plug this value into the equation:
v = u + at
v = 0 + (5.00 m/s^2) * (26.8 s)
v = 134 m/s
Therefore, the speed vTO of the plane as it takes off is 134 m/s.