Posted by JD on Tuesday, September 10, 2013 at 11:38pm.
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?

physics  Anonymous, Wednesday, September 11, 2013 at 1:08am
time*acceleration=134m/s

physics  Anthony, Saturday, September 14, 2013 at 7:58pm
134 m/s a=vt a=26.83s•5.00m/s^2

physics  Amal, Wednesday, May 27, 2015 at 6:21pm
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)
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