The landing speed of an aircraft is 54 m/s. If the maximum deceleration is 3m/s2 the minimum length of runway required is
- 360m.
- 486m.
- 162m.
Explanation
Using v squared = u squared minus 2as, with final speed zero, the stopping distance is v squared / (2a) = 54 squared / (2 x 3) = 2916 / 6 = 486 m. This is the minimum runway needed to bring the aircraft to rest at the stated maximum deceleration.
visco421 asking:
solve pls
Community Comments (2)
G
gyorwarth Posts: 2 06.06.2015 / 00:45
Hi, Hope this helps.
Using the equation of motion:
V^2=U^2+2as
Where:
V= initial velocity
U= final velocity (rest)
a= acceleration (or retardation)
s= displacement (length of runway required)
And substituting the question values into the equation we get:
54^2 = 0^2 + 2 x 3 x s
54 x 54 = 0 x 0 + 2 x 3 x s
2916 = 0 + 6s
2916 = 6s
2916 / 6 = s
496 = s
Using the equation of motion:
V^2=U^2+2as
Where:
V= initial velocity
U= final velocity (rest)
a= acceleration (or retardation)
s= displacement (length of runway required)
And substituting the question values into the equation we get:
54^2 = 0^2 + 2 x 3 x s
54 x 54 = 0 x 0 + 2 x 3 x s
2916 = 0 + 6s
2916 = 6s
2916 / 6 = s
496 = s
J
Jordi Posts: 198 12.06.2020 / 11:30
gyorwarth has a mistake at the end.It is 486 not 496.
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