An airplane has a velocity of 185 miles per hour when it lands, how long must the runway be if the plane decellerates at 2.5 meters per second
Vf^2=Vi^2+2ad
solve for d. a=-2.4m/s^2
To find the length of the runway required for the airplane to land, we need to convert the units of velocity and acceleration to a common unit.
First, let's convert the velocity from miles per hour to meters per second:
1 mile = 1609.34 meters (approximately)
1 hour = 3600 seconds
185 miles/hour * 1609.34 meters/mile * (1/3600) hours/second ≈ 82.64 meters/second
Now that we have the velocity of the plane in meters per second, which is 82.64 m/s, we can use the equation of motion to find the runway length.
The equation of motion relating the initial velocity (u), final velocity (v), acceleration (a), and displacement (s) is:
v^2 = u^2 + 2as
where:
v = final velocity = 0 m/s (since the plane comes to rest)
u = initial velocity = 82.64 m/s
a = acceleration = -2.5 m/s^2 (negative because it is decelerating)
s = displacement (runway length)
Plugging in the values, we get:
(0)^2 = (82.64)^2 + 2(-2.5)s
Simplifying the equation, we have:
0 = 6829.14 - 5s
Rearranging the equation:
5s = 6829.14
Dividing both sides by 5:
s ≈ 1365.83 meters
Therefore, the length of the runway must be approximately 1365.83 meters for the plane to come to a stop when decelerating at 2.5 meters per second.