A plane with constant acceleration until the half of the runway reaches the 80% of the velocity required for the take off.What fractional part of the track does it reach the take-off speed?

Let s1 = distance covered at halfway stage

Let s2 = distance still to be covered for takeoff

Let v1= velocity at halfway stage

let v2 = take off velocity

We have

v1² = 2as1 (1)

v2² -v1² = 2as2

v2² -(0.8v2)² = 2as2 (remember v1 was 80% of takeoff speed)

0.36 v2² = 2as2 (2)

Dividing (2) by (1)

0.36v2²/v1² =s2/s1

O.36 * (1.25)² =s2/s1 (remember v1 was 80% of takeoff speed)

s2/s1=0.5625

In other words 56.25 % of the distance covered so far is needed for takeoff

or 78.1% of the entire runway

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