Question

A 747 requires a takeoff speed of 270 meters per second if it is to get off the

ground. Calculate the acceleration required for the 747 that begins from rest
to takeoff on a 1000 meter long runway.

Answer 1

a = 36.45[m/s²]

Step-by-step explanation:

We can calculate the acceleration value by means of the following expression of kinematics.

`v_(f)^(2) =v_(o)^(2) +2*a*x`

where:

vf = final velocity = 270 [m/s]

Vo = initial velocity = 0 (begins from rest)

a = acceleration [m/s]²

x = distance required by the plane = 1000 [m]

`(270)^(2) =0+2*a*1000\\72900=2000*a\\a=36.45 [m/s^(2) ]`