$A$ motor car slows down from $72 \ km \ h^{-1}$ to $36 \ km \ h^{-1}$ over a distance of $25 \ m$. If the brakes are applied with the same force,calculate $(i)$ total time in which the car comes to rest,and $(ii)$ total distance travelled by it until it comes to rest.

(N/A) Case $1$: Initial velocity $u = 72 \ km \ h^{-1} = 20 \ m \ s^{-1}$; Final velocity $v = 36 \ km \ h^{-1} = 10 \ m \ s^{-1}$; Distance $S = 25 \ m$.
Using the equation of motion $v^{2} - u^{2} = 2aS$:
$(10)^{2} - (20)^{2} = 2 \times a \times 25$
$100 - 400 = 50a$
$-300 = 50a$
$a = -6 \ m \ s^{-2}$.
Case $2$: The car starts from $u = 72 \ km \ h^{-1} = 20 \ m \ s^{-1}$ and comes to rest $(v = 0)$ with the same retardation $a = -6 \ m \ s^{-2}$.
$(i)$ To find total time $t$,use $v = u + at$:
$0 = 20 + (-6)t$
$6t = 20$
$t = 3.33 \ s$.
$(ii)$ To find total distance $S$,use $v^{2} - u^{2} = 2aS$:
$0^{2} - (20)^{2} = 2 \times (-6) \times S$
$-400 = -12S$
$S = 400 / 12 = 33.33 \ m$.