A lorry and a car of mass ratio 4:1 are moving with KE in the ratio 3:2 on a horizontal road. Now brakes are applied and braking forces produced are in the ratio 1:2. Then the ratio of stopping timing of lorry and car is:

Mass of lorry and a car
$\frac{{\mathrm{m}}_{\mathrm{l}}}{{\mathrm{m}}_{\mathrm{c}}}=\frac{4}{1}$..........................(1)
The ratio of KE:

$\frac{{\mathrm{KE}}_{\mathrm{l}}}{{\mathrm{KE}}_{\mathrm{c}}}=\frac{3}{2}$…………………….(2)
And the ratio of breaking forces applied:

$\frac{{\mathrm{F}}_{\mathrm{l}}}{{\mathrm{F}}_{\mathrm{c}}}=\frac{1}{2}$................(3)
Now, let the initial velocities of the lorry and the car be u_l and u_c respectively.
We know that the kinetic energy is given by the expression
$\mathrm{KE}=\frac{1}{2}{\mathrm{mv}}^2$
So the initial kinetic energy of the lorry becomes
${\mathrm{KE}}_l=\frac{1}{2}{m}_l{u_l}^2$.....................(4)
And the kinetic energy of the car becomes
${\mathrm{KE}}_{\mathrm{c}}=\frac{1}{2}{\mathrm{m}}_{\mathrm{c}}{{\mathrm{u}}_{\mathrm{c}}}^2$..........................(5)
Putting (4) and (5) in (2) we get
$\frac{{\mathrm{KE}}_l=\frac{1}{2}{m}_l{u_l}^2}{{\mathrm{KE}}_c=\frac{1}{2}{m}_c{u_c}^2}=\frac{3}{2}$
Putting (1) in the above equation, we get

$\frac{4{u_l}^2}{1{u_c}^2}=\frac{3}{2}$

$\frac{u_l}{u_c}=\frac{\sqrt{3}}{\sqrt{8}}$

Also, from Newton’s second law of motion, we know that
$\mathrm{F} = \mathrm{ma}$
Therefore, if the retardations of the lorry and the car are a_l and a_c respectively, then their breaking forces are given by
$\mathrm{F}{ }_l= m_l{a}_l$..................(7)
${\mathrm{F}}_{\mathrm{c}} = m_c{a}_c$.....(8)
Putting (7) and (8) in (3) we get
$\frac{m_l{a}_l}{m_c{a}_c}=\frac{1}{2}$
Putting (1) in the above equation, we get
$\frac{{\mathrm{a}}_{\mathrm{l}}}{{\mathrm{a}}_{\mathrm{c}}}=\frac{1}{8}$.....................(9)
Now, from the second kinematic equation of motion, we have
$\mathrm{v} = \mathrm{u} + \mathrm{at}$
For the case of retardation, a = -a
 $\mathrm{v} = \mathrm{u} - \mathrm{at}$
Since both the lorry and the car have been stopped, we substitute v = 0 in the above to get
$0 = \mathrm{u} - \mathrm{at}$
$\Rightarrow \mathrm{t} =\frac{ \mathrm{u}}{\mathrm{a}}$
Therefore, the stopping time for the lorry and the car are respectively given by
$\Rightarrow {\mathrm{t}}_{\mathrm{l}} =\frac{ {\mathrm{u}}_{\mathrm{l}}}{{\mathrm{a}}_{\mathrm{l}}}$.........................(10)
$\Rightarrow {\mathrm{t}}_c =\frac{ {\mathrm{u}}_c}{{\mathrm{a}}_c}$..............(11)
Dividing (10) by (11) we get
$\Rightarrow \frac{{\mathrm{t}}_{\mathrm{l}} }{{\mathrm{t}}_{\mathrm{c}}}=\frac{{\displaystyle \raisebox{1ex}{${\mathrm{u}}_{\mathrm{l}}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{u}}_{\mathrm{c}}$}\right.} }{{\displaystyle \raisebox{1ex}{${\mathrm{a}}_{\mathrm{l}}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{a}}_{\mathrm{c}}$}\right.}}$
Substituting (9) and (6) in the above equation, we finally get

$\frac{t_l}{t_c}=8\times \frac{\sqrt{3}}{2\sqrt{2}}=2\sqrt{6}$

Thus, the ratio of the stopping time of the lorry and the car is equal to $2\sqrt{6}:1$.