an air craft has a takeoff velocity of 180km/h. what length of runway is needed? the accelaration of aircraft along the runway is 2.5 m/s^2
V^2=2as.
Convert 180km/hr into m/s.
180000/3600=50m/s.
Then put in-
v^2=2as
(50)^2=2*(2.5m/s)^2*s
2500=5*s
s=500m
2505
To find the length of runway needed for an aircraft with a given takeoff velocity and acceleration, we can use the kinematic equation:
v² = u² + 2as
Where:
v = final velocity (takeoff velocity)
u = initial velocity (0 in this case as the aircraft starts from rest)
a = acceleration
s = distance (length of runway)
In this case, we have:
v = 180 km/h = 180 * (1000/3600) m/s = 50 m/s
u = 0 m/s
a = 2.5 m/s²
Plugging these values into the kinematic equation, we have:
(50)² = (0)² + 2 * 2.5 * s
Simplifying,
2500 = 5s
Dividing both sides by 5,
s = 500 m
Therefore, the length of runway needed for this aircraft is 500 meters.
v^2 = 2as
convert km/h to m/s and plug in the numbers to find s.