An aircraft is at rest at one end of a runway which is 2.2 km long.

Question 33

An aircraft is at rest at one end of a runway which is 2.2 km long. The aircraft accelerates along the runway with an acceleration of 2.5 m s^-2 until it reaches its take-off speed of 75 m s^-1. Calculate

(i) the time taken to reach take-off speed

(ii) the distance travelled in this time

(iii) Just as the aircraft reaches take-off speed a warning light turns on. The maximum possible deceleration of the aircraft is 4.0 m s^-2 and 2.5 s elapses before the pilot takes any action, during which times the aircraft continues at its take off speed. Determine whether or not the aircraft can be brought to rest in the remaining length of runway.

Reference: Cambridge International AS and A Level Physics Coursebook – 2^nd edition – by David Sang, Graham Jones, Gurinder Chadha and Richards Woodside

Solution:

(i)

Initial speed, u = 0 m s^-1 (aircraft is at rest)

Acceleration, a = 2.5 m s^-2

Final speed, v = 75 m s^-1 (take-off speed)

a = (v - u) / t

Time taken to reach take-off speed, t = (v – u) / a = (75 – 0) / 2.5 = 30 s

(ii)

Consider the equation for uniformly accelerated motion:

s = ut + ½ at²

Distance travelled during this time, s = (0×30) + (½ ×2.5×30²) = 1125 m

Alternatively,

Consider the equation for uniformly accelerated motion:

v² = u² + 2as

75² = 0² + (2×2.5×s)

Distance travelled during this time, s = 75² / (2×2.5) = 1125 m

(iii)

Remaining length of runway when warning light turns on = 2200 – 1125 = 1075 m

Distance travelled during the 2.5 s before the pilot takes any action:

Speed (= distance / time) = 75 m s^-1

Distance = Speed × Time = 75 × 2.5 = 187.5 m

Remaining length of runway for deceleration = 1075 – 187.5 = 887.5 m

Consider the deceleration to be maximum (= 4 m s^-2) [acceleration, a = – 4 m s^-2]

Initial speed, u = 75 m s^-1 and final speed, v = 0 m s^-1 (aircraft brought to rest)

Let’s try to calculate the distance s required to bring the aircraft to rest and compare this with the remaining length of runway.

Consider the equation for uniformly accelerated motion:

v² = u² + 2as

0² = 75² + 2×-4×s

s = 75² / 8 = 703.125 m

Since the distance required to bring the aircraft to rest (703.125 m) is less than the remaining length of runway (887.5 m), the aircraft can be brought to rest in the remaining length of runway.