An engineer messed design a runway to accommodate airplanes that must reach a speed of 71 m/s before they can take off the planes are capable of accelerating at a rate of 1.6 m/s person per second the minimum length of the runway must be _______
Vf^2 = Vo^2 + 2a*L.
Vf = 71 m/s.
Vo = 0.
a = 1.6 m/s^2.
L = ?
To find the minimum length of the runway, we can use the equation of motion:
v² = u² + 2as
where:
v = final velocity of the airplane (71 m/s)
u = initial velocity of the airplane (0 m/s, since the airplane starts from rest)
a = acceleration of the airplane (1.6 m/s²)
s = distance traveled (runway length)
Rearranging the equation to solve for s, we have:
s = (v² - u²) / (2a)
Now, let's substitute the given values into the equation:
s = (71² - 0²) / (2 * 1.6)
s = (5041 - 0) / 3.2
s = 5041 / 3.2
s ≈ 1575.31 meters
Therefore, the minimum length of the runway must be approximately 1575.31 meters.