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

Answer:

a = 36.45[m/s²]

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]

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