The motors of an electric train can give it an acceleration of 1 m/s² and brakes can give a negative acceleration of 3 m/s². The shortest time in which the train -can make a trip between two stations .1215 m apart is

The velocity time graph is shown in fig. (a)
From the given problem,
$\frac{\mathrm{V}}{{\mathrm{t}}_1}=1 \mathrm{or} {\mathrm{t}}_1=\mathrm{V}$
similarly, $\frac{\mathrm{V}}{{\mathrm{t}}_2}=3 \mathrm{or} {\mathrm{t}}_2=(\mathrm{v}/3)$
Here $1215=\frac{1}{2}\mathrm{v}\left({\mathrm{t}}_1+{\mathrm{t}}_2\right)=\frac{1}{2}\mathrm{v}\left(\mathrm{v}+\frac{\mathrm{v}}{3}\right)$
$1215=\frac{2{\mathrm{v}}^2}{3} \mathrm{or} {\mathrm{v}}^2=\frac{3645}{2}$
$\mathrm{V}=\sqrt{\left(\frac{3645}{2}\right)}=\sqrt{(1822⋠5)}=42⋠69$
Now ${\mathrm{t}}_1+{\mathrm{t}}_2=\mathrm{v}+\frac{\mathrm{V}}{3}=\frac{4×42⋠69}{3}=56⋠92\sec$