Question

An airplane must reach a speed of 322 km/h to take off. If the runway is 500 m long, what is the minimum value of the acceleration that will allow the airplane to take off successfully?

Answer 1

The relation between the trace of a body under acceleration with the last one is given by a formula:

$T = V_0t+\frac{at^2}{2}$ T - a trace, t - a time, $V_0, a$ - the initial velocity and the acceleration

$V_0=0\implies T = \frac{at^2}{2}$(1)

The velocity $V$ of the body in time $t$:

$V = V_0 + at= at\implies t = \frac{V}{a}$(2)

Sibstitute(2) into (1):

$T = \frac{V^2}{2a}\implies a = \frac{V^2}{2T}$

$a = \frac{322*322}{2*0.5} = 103684\frac{km}{h^2}$