A Boeing 787 is initially moving down the runway at 6.0 m/s preparing for takeoff. The pilot pulls on the throttle so that the engines give the plane a constant acceleration of 1.9 m/s2. The plane then travels a distance of 1700 m down the runway before lifting off. How long does it take from the application of the acceleration until the plane lifts off, becoming airborne

39.26 s

Explanation:

From the question given above, the following data were obtainedb

Initial velocity (u) = 6 m/s

Acceleration (a) = 1.9 m/s²

Distance travelled (s) = 1700 m

Time (t) =?

Next, we shall determine the final velocity of the plane. This can be obtained as follow:

Initial velocity (u) = 6 m/s

Acceleration (a) = 1.9 m/s²

Distance travelled (s) = 1700 m

Final velocity (v) =?

v² = u² + 2as

v² = 6² + (2 × 1.9 × 1700)

v² = 36 + 6460

v² = 6496

Take the square root of both side

v = √6496

v = 80.6 m/s

Finally, we shall determine the time taken before the plane lifts off. This can be obtained as follow:

Initial velocity (u) = 6 m/s

Acceleration (a) = 1.9 m/s²

Final velocity (v) = 80.6 m/s

Time (t) =?

v = u + at

80.6 = 6 + 1.9t

Collect like terms

80.6 – 6 = 1.9t

74.6 = 1.9t

Divide both side by 1.9

t = 74.6 / 1.9

t = 39.26 s

This, it will take 39.26 s before the plane lifts off.