A person is running at a maximum speed of $4{\text{ms}}^{-1}$ to catch a train. When he is 6m from the door of the compartment, the train starts to leave the station at a constant acceleration of $1{\text{ms}}^{-2}.$ The time taken by the person to catch the train is

man${\text{v}}_{\text{m}}=4{\text{ms}}^{-1},\text{t}=\text{t}$

Train$\text{u}=0{\text{ms}}^{-1},\text{a}=1{\text{ms}}^{-1},{\text{s}}_{\text{train}}=?$

Distance travelled by man $\left({\text{s}}_{\text{man}}\right)=6+$ distance traveled by train

${\text{s}}_{\text{man}}=6+{\text{s}}_{\text{train}}$

${\text{v}}_{\text{m}}\times \text{t}=6+\text{ut}+\frac{1}{2}{\text{at}}^{\text{2}}$

$4\text{t}=6+\text{0}+\frac{1}{2}\left(1\right){\text{t}}^{\text{2}}$

$4\text{t}=6+\frac{{\text{t}}^{\text{2}}}{2}$

$4\text{t}=\frac{\text{12}+{\text{t}}^{\text{2}}}{2}$

${\text{t}}^{\text{2}}+8\text{t}+12=0$

${\text{t}}^{\text{2}}-6\text{t}-2\text{t}+12=0$

$\text{t}\left(\text{t}-6\right)-2\left(\text{t}-6\right)=0$

$\text{t}=\text{2}\left(\text{or}\right)\text{t}=6$