The speed of a body moving with uniform acceleration is u. This speed is doubled while covering a distance S. When it covers an additional distance S, its speed would become

Options

A. `sqrt3u`
B. `sqrt5u`
C. `sqrt11u`
D. `sqrt7u`

Explanation

To solve the problem step by step, we will use the equations of motion under uniform acceleration.

### Step 1: Understand the initial conditions
We are given that the initial speed of the body is \( u \) and it doubles its speed while covering a distance \( S \).

### Step 2: Apply the kinematic equation
Using the third equation of motion:
\[ v^2 = u^2 + 2as \]
where:
- \( v \) is the final velocity,
- \( u \) is the initial velocity,
- \( a \) is the acceleration,
- \( s \) is the distance covered.

In this case, when the body covers the first distance \( S \), its final speed becomes \( 2u \). Thus, we can write:
\[ (2u)^2 = u^2 + 2aS \]
This simplifies to:
\[ 4u^2 = u^2 + 2aS \]

### Step 3: Rearrange to find acceleration
Rearranging the equation gives:
\[ 4u^2 - u^2 = 2aS \]
\[ 3u^2 = 2aS \]
From this, we can express acceleration \( a \):
\[ a = \frac{3u^2}{2S} \]

### Step 4: Analyze the second distance
Now, we need to find the speed after covering an additional distance \( S \). The total distance covered now is \( 2S \) and the initial speed for this part of the journey is \( 2u \).

### Step 5: Use the kinematic equation again
Using the same kinematic equation for the second part of the journey:
\[ v^2 = (2u)^2 + 2a(2S) \]
Substituting \( a \) from the previous step:
\[ v^2 = 4u^2 + 2\left(\frac{3u^2}{2S}\right)(2S) \]
This simplifies to:
\[ v^2 = 4u^2 + 3u^2 \]
\[ v^2 = 7u^2 \]

### Step 6: Solve for final velocity
Taking the square root of both sides gives:
\[ v = \sqrt{7}u \]

### Final Answer
The final speed after covering an additional distance \( S \) is:
\[ v = \sqrt{7}u \]