For an aroplane to take off it accelerates according to graph shown in figure and takes 1,2 s to take off from the position. The distance travelled by the aroplane is : 

From O to A, the acceleration is varying linearly with time, hence
$\mathrm{a}−0=\frac{5}{6}(\mathrm{t}−0) \mathrm{or} \mathrm{a}=\frac{5}{6}\mathrm{t}$
Now $\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{5}{6}\mathrm{t} \mathrm{or} {∫}_0^{{\mathrm{v}}_1} \mathrm{dv}=\frac{5}{6}{∫}_0^{\mathrm{t}} \mathrm{tdt}$
$∴{\mathrm{v}}_1=\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{5}{12}{\mathrm{t}}^2$
Further, ${∫}_0^{{\mathrm{x}}_1} \mathrm{dx}=\frac{5}{12}{∫}_0^{\mathrm{t}} {\mathrm{t}}^2\mathrm{dt}$
${\mathrm{x}}_1=\frac{5}{12}{\left[\frac{{\mathrm{t}}^3}{3}\right]}_0^{\mathrm{t}}=\frac{5}{36}{\mathrm{t}}^3$
or ${\mathrm{x}}_1=\frac{5}{36}(6{)}^3=30\mathrm{m}$
From A to B, the acceleration is constant. It's value is 5 m/s². So, the velocity at Point A is 
${\mathrm{v}}_1=\frac{5}{12}\left(36\right)=15\mathrm{m}/\mathrm{s}$
or ${\mathrm{x}}_2={\mathrm{v}}_1\left(6\right)+\frac{1}{2}\left(5\right)(6{)}^2=180\mathrm{m}$
Total distance travelled is given by 
x=30+180=210m