V = 117km/h = 117000m/3600s. = 32.5 m/s.
V^2 = Vo^2 + 2a*L = 32.5^2
0 + 4.8*L = 1056.25
L = 1056.25/4.8 = 220 m.
V = Vo + a*t = 32.5 m/s.
0 + 2.4*
A small, single engine airplane is about to take off. The airplane becomes airborne, when its speed reaches 117.0 km/h. The conditions at the airport are ideal, there is no wind. When the engine is running at its full power, the acceleration of the airplane is 2.40 m/s2. What is the minimum required length of the runway?
I thought you used the equation: Vf^2=Vi^2+2ad where Vi is 0 but I can't seem to get the right answer. Can someone help please?!?
2 answers
V = 117km/h = 117000m/3600s. = 32.5 m/s.
V^2 = Vo^2 + 2a*L = 32.5^2
0 + 4.8*L = 1056.25
L = 1056.25/4.8 = 220 m.
V^2 = Vo^2 + 2a*L = 32.5^2
0 + 4.8*L = 1056.25
L = 1056.25/4.8 = 220 m.
9 answers