Answer:
The answer is 3.33m
Explanation:
The acceleration "a" is constant.
Acceleration is the variation of velocity over time,
[tex]\frac{dv}{dt} = a[/tex].
solving the last equation
[tex]\int_{v_0}^v dv = a\int_0^t dt \rightarrow v-v_0 = at[/tex],
where [tex]v_0=0[/tex] because the airplane starts from rest.
Once again, velocity is the variation of distance over time.
[tex]\frac{dx}{dt} = at \rightarrow \int_{x_0}^x dx = a\int_0^t t\ dt[/tex]
then
[tex]x- x_0 = \frac{1}{2}at^2[/tex]
where [tex]x_0=0[/tex] if we consider the end of the runway as the initial point (this step is for simplicity but you can let it expressed, it's going to cancel anyway).
If [tex]x=1.11\ m[/tex] at [tex]t=1s[/tex], then
[tex]a = \frac{2x}{t^2} = 2.22\ m/s^2[/tex]
and the final expression for the distance is
[tex]x = 1.11 t^2[/tex].
If t = 2s, x = 4.44 m. Which means thad the additional distance is
[tex]x(2s) - x(1s) = 4.44 - 1.11 = 3.33\ m[/tex]
