What percentage of the takeoff velocity did the plane gain when it reached the midpoint of the runway?

To take off from the ground, an airplane must reach a sufficiently high speed. The velocity required for the takeoff, the takeoff velocity, depends on several factors, including the weight of the aircraft and the wind velocity.



What percentage of the takeoff velocity did the plane gain when it reached the midpoint of the runway?



Im not sure how to set this problem up.

I know it takes 26.83 s. to take off

i know the speed of plane as it takes off is 134 m/sWhat percentage of the takeoff velocity did the plane gain when it reached the midpoint of the runway?If you assume (1) constant acceleration, and (2) takeoff velocity is reached at the end of the runway, then the solution is very simple.

From the basic distance-acceleration-time relationship, d = at^2/2 we find t = sqrt(2d/a), and we can easily derive v = at = sqrt(2da).

So given that v is proportional to sqrt(d), if d1/d2 = 1/2, v1/v2 = sqrt(d1/d2) = sqrt(1/2) = 0.707107 or 70.7107%.What percentage of the takeoff velocity did the plane gain when it reached the midpoint of the runway?What percentage of the takeoff velocity did the plane gain when it reached the midpoint of the runway?



(Final velocity)^2 – (Initial velocity)^2 = 2 * acceleration * distance



vi = initial velocity = 0

vf = take off velocity

a = acceleration

d = length of runway



vf^2 – 0 = 2 * a * d

vf^2 = 2 * a * d



vf = (2 * a * d)^0.5

In the equation above, vf is the takeoff velocity of a plane accelerating at “a” m/s^2 for “d” meters.



The velocity of the plane at d/2 meters = (2 * a * d/2)^0.5





% of the takeoff velocity = [(2 * a * d/2)^0.5] ÷ [(2 * a * d)^0.5] * 100%



The (2 * a) will cancel

% of the takeoff velocity = [(d/2)^0.5 ÷ d^0.5] * 100%



[(d/2)^0.5 ÷ d^0.5] = (d/2 ÷ d)^0.5 = (?)^0.5 = 0.7071



% of the takeoff velocity = 70.7%





Using your numbers!!



I know it takes 26.83 s. to take off

i know the speed of plane as it takes off is 134 m/s



The acceleration = 134 ÷ 26.83 = 5 m/s^2



Distance = ? * acceleration * time^2

Distance = ? * 5 * 26.83^2 = 1800 m



What is the velocity at ? * 1800 m = 900 m?

What is the time at 900 m



900 = 2.5 * t^2

t^2 = 900 ÷ 2.5 = 360

t = 360^0.5 = 18.97 seconds



final velocity = acceleration * time



final velocity = 5 * 18.97 = 94.85 m/s



% of the takeoff velocity = (94.85 ÷ 134) * 100%



% of the takeoff velocity = 70.8 %