An aircraft has 14000 lbs of net thrust weighs 36000 lbs and

An aircraft has 14,000 lbs of net thrust, weighs 36,000 lbs, and Vr is 160 kts. How many feet down the runway would the aircraft be at Vr (rotation speed). The aircraft starts from a standing start 135 feet down the runway.

Solution

We have the net thrust acting on the aircraft of mass 36000 lbs.

Therefore the acceleration the aircraft would suffer would be given as: Force / Mass

or, Acceleration = 14000 x 32.2 / 36000 = 14 x 32.2 / 36 = 12.52 ft/s²

Now we know that 1 knot is equal to 1.68781 ft/sec

So the distance required to achieve 160 knots or 270.05 ft/sec would be given as:

V² = u² + 2aS

or, 270.05² = 2 x 12.52 x S

or, S = 2912.420 ft

Therefore the aircraft will need to travel 2912.42 feet to achieve Vr.