**Problem statement**
A body moving with uniform retardation covers 3km before its speed is reduced to half of its initial value. It comes to rest in another distance of ________.
Options:
1. 1 km
2. 2 km
3. 3 km
4. 1/2 km

**Solution**
Given, the body moves with uniform retardation which means the acceleration is negative and constant.
Let's assume the initial velocity of the body as u and the final velocity as v.
We know that the average velocity of the body during the motion is given by (u+v)/2.
Also, we know that the distance covered by the body is given by the formula, s = (u+v)t/2, where t is the time taken.
Since the body comes to rest, the final velocity v = 0.

**Finding time**
Using the above formulas, we can find the time taken for the body to reach half of its initial velocity.
Let the time taken to reach half of the initial velocity be t1.
s = (u+v)t/2
3 = (u+v)t1/2
=> t1 = 6/(u+v)

**Finding distance**
Now, let's find the distance covered by the body when its speed is reduced to half of its initial value.
Using the formula, v = u + at, where a is the retardation, we can find the time taken for the body to reach half of its initial velocity.
v = u/2
u + at = u/2
=> t = u/2a
Using the formula, s = ut + (1/2)at^2, we can find the distance covered by the body from the initial velocity to half of the initial velocity.
s1 = ut + (1/2)at^2
s1 = u(u/2a) + (1/2)a(u/2a)^2
s1 = u^2/4a
Using the formula, s = (u+v)t/2, we can find the distance covered by the body from half of the initial velocity to rest.
s2 = (u/2+0)t1/2
s2 = u/4t1
substituting t1 = 6/(u+v)
s2 = 3u/(u+v)^2

Total distance covered by the body = s1 + s2
= u^2/4a + 3u/(u+v)^2

**Substituting values**
Given, the speed of the body is reduced to half of the initial velocity after covering 3 km.
3 = (u+v)t1/2
=> t1 = 6/(u+v)
v = u/2
=> u+v = 3u/2
=> (u+v)^2 = 9u^2/4
Substituting these values in the total distance formula,
Total distance covered by the body = u^2/4a + 3u/(u+v)^2
= u^2/4a + 3u/(9u^2/4)
= u^2/4a + 4/3u

Given, the body covers 3 km before reducing its speed to half of its initial velocity.
3 = (u+v)t1/2
=> t1 = 6/(u