Question

An airplane travels 330 m down the runway before taking off. Assuming that it has constant acceleration, if it starts from rest and becomes airborne in 9.00 s, how fast (in m/s) is it moving

Answer 1

Under constant acceleration, we have average velocity $\bar v$ over a time interval of duration $\Delta t$ satisfying

$\bar v=\dfrac{v+v_0}2=\dfrac{x-x_0}{\Delta t}$

where $x_0,v_0$ are the plane's initial position and velocity (respectively) and $x,v$ are the plane's position and velocity (respectively) after 9.00 seconds. So

$\dfrac{v+0\,\frac{\mathrm m}{\mathrm s}}2=\dfrac{330\,\mathrm m-0\,\mathrm m}{9.00\,\mathrm s}\implies v=73.3\,\dfrac{\mathrm m}{\mathrm s}$