A jumbo jet must reach a speed of 360 km/h on the runway for takeoff. What is the least constant acceleration needed for takeoff from a 1.80 km runway?

1 Answer

2.777\ m/s^2

Explanation:

The least acceleration a of jumbo jet starting from rest u=0 to reach a velocity of v=360\ \text{km/hr}=100\ m/s running on runway of s=1.8\ km=1800\ m to take off is given by third equation of motion

v^2=u^2+2as

100^2=0^2+2a(1800)

a=\frac{10000}{3600}

a=2.7777\ m/s^2