A body moving with a uniform retardation covers 3 km before its speed is reduced to half of its initial value. It comes to rest in another distance of
$\left( a \right)$ 1 Km
$\left( b \right)$ 2 Km
$\left( c \right)$ 3 Km
$\left( d \right)$ (1/2) Km

Hint: In this particular question use the concept that in the question time is not present so we use the equations of motion in which time is not present this will make task easy so we use third equation of motion which is given as (${v^2} = {u^2} + 2as$) where symbols have their usual meaning so use this concept to reach the solution of the question.

Complete step by step solution:
A body moving with a uniform retardation covers 3 km before its speed is reduced to half of its initial value.
Let initial velocity be u Km/hr.
So its final velocity is half of initial velocity, so final velocity is, v = $\dfrac{u}{2}$ Km/hr.
During this interval the body covers 3 Km.
Therefore, s = 3 Km.
Now according to third equation of motion we have,
${v^2} = {u^2} + 2as$, where, v = final velocity, u = initial velocity, a = retardation and s = distance covered.
Now substitute the values we have,
$ \Rightarrow {\left( {\dfrac{u}{2}} \right)^2} = {u^2} + 2a\left( 3 \right)$
Now simplify we have,
$ \Rightarrow 6a = \dfrac{{{u^2}}}{4} - {u^2} = \dfrac{{ - 3{u^2}}}{4}$
$ \Rightarrow a = \dfrac{{ - {u^2}}}{8}$, (‘-’ sign indicates retardation)
Now we have to find the distance covered by the body so that it comes to res.
So for this the body initial velocity becomes u’ = (u/2) and final velocity, v’ = 0 and retardation is uniform so it remains the same.
Let the distance body covers be s’ Km.
So again by third equation of motion we have,
$ \Rightarrow {\left( {v'} \right)^2} = {\left( {u'} \right)^2} + 2a\left( {s'} \right)$
Now substitute the values we have,
$ \Rightarrow {\left( 0 \right)^2} = {\left( {\dfrac{u}{2}} \right)^2} + 2\left( {\dfrac{{ - {u^2}}}{8}} \right)\left( {s'} \right)$
$ \Rightarrow 0 = \dfrac{{{u^2}}}{4} - \dfrac{{{u^2}}}{4}\left( {s'} \right)$
$ \Rightarrow \dfrac{{{u^2}}}{4} = \dfrac{{{u^2}}}{4}\left( {s'} \right)$
$ \Rightarrow s' = 1$ Km.
So the body comes to rest at another distance of 1 Km.
Hence option (A) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the equations of motion i.e. first equation of motion is $v = u + at$, second equation of motion is $s = ut + \dfrac{1}{2}a{t^2}$ and third equation of motion is stated above so just substitute the values according to question in the third equation of motion as above and simplify we will get the required answer.