Question

Calculate the point of no return for an airport runway of 1.60 mi in length if a jet plane can accelerate at 10.1 ft/s2 and decelerate at 7.21 ft/s2. The point of no return occurs when the pilot can no longer abort the takeoff without running out of runway.
(a) What length of time is available from the start of the motion in which to decide on a course of action?

Answer 1

Answer:

t = 26.39 s

Explanation:

given,

Length of runway = 1.60 mi = 8448 ft

acceleration of the jet = 10.1 ft/s²

deceleration of the jet = 7.21 ft/s²

Let x be the distance of no return

using equation of motion

v² = u² + 2 a s

v² = 0 + 2 × 10.1 × x

v = √(20.2 x)

now, after reaching that Point the deceleration of the plane start

using equation of motion

final velocity of the first case will be the initial velocity of the second case

v² = u² + 2 a s

0 = 20.2 x - 2 x 7.21 x(8448-x)

34.62 x = 121820.16

x = 3518.78 ft

time taken to reach no return point

$x = ut+\dfrac{1}{2}at^2$

$3518.78 =\dfrac{1}{2}\times 10.1\times t^2$

    t² = 696.788

    t = 26.39 s

time taken to reach the no return point is equal to 26.39 s.