Question

An aircraft has a liftoff speed of 33 m/s. What minimum constant acceleration does this require if the aircraft is to be airborne after a take-off run of 240 m

Answer 1

2.27 m/s²

Step-by-step explanation:

Recall that one of the equations of motion can be expressed as

v² = u² + 2as

where

v = final velocity = takeoff speed = 33m/s

u = initial velocity = rest = 0 m/s

a = acceleration ( we are asked to find this)

s = distance traveled = length of takeoff run = 240m

substituting the known values into the equation above,

v² = u² + 2as

33² = 0² + 2a(240)

1089= 480a

a = 1089/480

a = 2.27 m/s²

Answer 2

2.27 m/s^2

EXPLANATION

We can use the general relationship from kinematics:

vf^2 = vi^2 + 2as

where:

s = distance;

a = acceleration;

vi = 0m/s is the initial velocity;

vf=33ms is the final velocity.

So:

33^2 = 0^2 + 2 ⋅ a ⋅ 240

a= 1089/480

= 2.27m/s^2