Thank you for reporting, we will resolve it shortly
Q.
A car moving with a velocity of $20\, ms^{-1}$ is stopped in a distance of $40\, m$. If the same car is travelling at double the velocity, the distance travelled by it for same retardation is
Initial velocity of car, $u =20 m / s$
The final velocity will be, $v =0$ as the car comes to rest after travelling distance of $40 m$ (which will be after application of brakes)
By third equation of motion we know,
$
\begin{array}{l}
v^{2}=u^{2}+2 \text { as } \\
0=(20)^{2}+2 \times a \times(40) \\
-80 a=400
\end{array}
$
Thus, $a=-5 m / s ^{2}$
Negative sign indicates that it is a retardation.
When same car is moving with double the velocity
Initial velocity, $u=2 \times 20=40 m / s$
Thus,
$
\begin{array}{l}
0=(40)^{2} \times 2 \times(-5) \times s \\
10 s=1600 \\
s=160 m
\end{array}
$
Hence, if the same car is travelling at double the velocity the distance travelled by it for same retardation is $160 m$