A body moving with uniform retardation covers 3 km before its speed is reduced to half of its initial value. It comes to rest in another distance of

Options

A. 1 km
B. 2 km
C. 3 km
D. `(1)/(2)` km

Answer

To solve the problem step by step, we will use the equations of motion under uniform acceleration (or retardation in this case).

### Step 1: Define the variables
Let:
- Initial velocity = \( v \)
- Final velocity after covering 3 km = \( \frac{v}{2} \)
- Distance covered before speed is halved = \( s_1 = 3 \, \text{km} = 3000 \, \text{m} \)
- Retardation = \( -a \)

### Step 2: Use the equation of motion
We will use the equation of motion:
\[ v^2 = u^2 + 2as \]
Where:
- \( v \) = final velocity
- \( u \) = initial velocity
- \( a \) = acceleration (negative for retardation)
- \( s \) = distance covered

Substituting the known values:
\[ \left(\frac{v}{2}\right)^2 = v^2 + 2(-a)(3000) \]

### Step 3: Simplify the equation
This simplifies to:
\[ \frac{v^2}{4} = v^2 - 6000a \]
Rearranging gives:
\[ 6000a = v^2 - \frac{v^2}{4} \]
\[ 6000a = \frac{3v^2}{4} \]
Thus, we can express \( a \) as:
\[ a = \frac{3v^2}{24000} = \frac{v^2}{8000} \]

### Step 4: Calculate the distance to come to rest
Now, we need to find the distance \( s_2 \) covered after the speed is halved until the body comes to rest. The initial velocity for this phase is \( \frac{v}{2} \) and the final velocity is \( 0 \).

Using the same equation of motion:
\[ 0 = \left(\frac{v}{2}\right)^2 + 2(-a)s_2 \]
Substituting for \( a \):
\[ 0 = \frac{v^2}{4} - 2\left(\frac{v^2}{8000}\right)s_2 \]

### Step 5: Solve for \( s_2 \)
Rearranging gives:
\[ 2\left(\frac{v^2}{8000}\right)s_2 = \frac{v^2}{4} \]
Dividing both sides by \( \frac{v^2}{8000} \):
\[ s_2 = \frac{\frac{v^2}{4}}{2\left(\frac{v^2}{8000}\right)} = \frac{8000}{8} = 1000 \, \text{m} = 1 \, \text{km} \]

### Final Answer
The body comes to rest after covering an additional distance of **1 km**.

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