Question

A plane starting at rest at the south end of a runway undergoes a constant acceleration of 1.6 m/s2 for a distance of 1400 m before takeoff.

(a) What is the plane's velocity at takeoff?

Answer 1

velocity at takeoff = 66.93 [m/s]

Step-by-step explanation:

For the solution we use the kinematic equation:

`V^(2)=Vi^(2)+2*a*(x-xi) \\ where:\\V=final velocity[m/s]\\Vi=initial velocity[m/s]\\a=acceleration[m/s^(2) ]\\x=rest position [m]\\xi=initial position [m]\\\\`

Now replacing the data

`V^(2) = 0 + 2*1.6*(1400-0)\\V=66.93[m/s]`