when an aircraft takes off, it accelerates until it reaches its takeoff speed V.In doing so it uses up a distance R of the runway, where R is proportional to the square of the of the takeoff speed. If V is measured in miles per hour and R is measured in feet, then 0.1639 is the constant of proportionality. If an aircraft has a takeoff speed of about 215 miles per hour, how much runway does it need?
R = .1639 V^2
R = .1639 * 215^2
To find the distance of runway needed for takeoff, we can use the given information that the distance R is proportional to the square of the takeoff speed V, with a constant of proportionality of 0.1639.
Let's denote the distance of runway needed as D. We can set up the proportionality equation as follows:
D / R = V^2 / (V0^2)
Where D is the distance of the runway needed, R is the constant of proportionality, V is the takeoff speed, and V0 is a reference takeoff speed.
Now, we can rearrange the equation to solve for D:
D = R * (V^2 / (V0^2))
Given that R = 0.1639 and V = 215 miles per hour, we need to determine the value of V0. In this case, we can use a reference takeoff speed of V0 = 1 mile per hour.
Substituting the values into the equation, we have:
D = 0.1639 * (215^2 / 1^2)
Calculating this expression:
D = 0.1639 * (46225 / 1)
D = 0.1639 * 46225
D ≈ 7563.6475
Therefore, the aircraft needs approximately 7,563.65 feet of runway for takeoff.