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 retardation.

### Given:
- Initial distance covered before speed is halved: \( S_1 = 3 \, \text{km} = 3000 \, \text{m} \)
- Initial speed: \( v \)
- Speed after covering 3 km: \( v_f = \frac{v}{2} \)
- Distance covered to come to rest: \( S_2 = x \, \text{km} \)

### Step 1: Use the first equation of motion for the first part of the motion
Using the equation:
\[ v_f^2 = v^2 + 2a S \]
For the first part of the motion, we have:
\[ \left(\frac{v}{2}\right)^2 = v^2 + 2(-a)(3000) \]
This can be rewritten as:
\[ \frac{v^2}{4} = v^2 - 6000a \]

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

### Step 3: Use the second equation of motion for the second part of the motion
For the second part of the motion, where the body comes to rest:
\[ 0 = \left(\frac{v}{2}\right)^2 + 2(-a)(x) \]
This can be rewritten as:
\[ 0 = \frac{v^2}{4} - 2ax \]

### Step 4: Rearranging the second equation
Rearranging gives:
\[ 2ax = \frac{v^2}{4} \]
Substituting \( a \) from Step 2 into this equation:
\[ 2\left(\frac{v^2}{8000}\right)x = \frac{v^2}{4} \]
This simplifies to:
\[ \frac{v^2}{4000}x = \frac{v^2}{4} \]

### Step 5: Canceling \( v^2 \) and solving for \( x \)
Assuming \( v^2 \neq 0 \), we can divide both sides by \( v^2 \):
\[ \frac{x}{4000} = \frac{1}{4} \]
Multiplying both sides by 4000 gives:
\[ x = 1000 \, \text{m} = 1 \, \text{km} \]

### Final Answer:
The distance the body covers to come to rest is \( x = 1 \, \text{km} \).

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