An airplane starts from rest on the runway. The engines exert a constant force of 78.0...
An airplane starts from rest on the runway. The engines exert a constant force of 78.0 kN on the body of the plane (mass 9.20 × 104 kg) during takeoff. How far down the runway does the plane reach its takeoff speed of 74.7 m/s?
Homework Answers
acceleration acts on plane
a = F/m = 78000 / ( 9.2* 10^4)
a = 0.8478 m/s^2
using 3rd equation of motion
v^2 = 2 a d
74.7^2 = 2* 0.8478* d
d = 74.7^2 / ( 2* 0.8478)
d = 5580.09 / 1.6956
d = 3290.82 m
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