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. `sqrt(3)u`
B. `sqrt(5)u`
C. `sqrt(11)u`
D. `sqrt(7)u`

Answer

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

### Step 1: Understand the problem
We start with a body moving with an initial speed \( u \) and it doubles its speed to \( 2u \) while covering a distance \( S \). We need to find the speed of the body after it covers an additional distance \( S \).

### Step 2: Apply the first case (from initial speed \( u \) to \( 2u \))
Using the third equation of motion:
\[ v^2 = u^2 + 2as \]
Here, \( v = 2u \), \( u = u \), and \( s = S \).
Substituting these values into the equation:
\[ (2u)^2 = u^2 + 2aS \]
This simplifies to:
\[ 4u^2 = u^2 + 2aS \]
Rearranging gives:
\[ 4u^2 - u^2 = 2aS \implies 3u^2 = 2aS \]
From this, we can express \( a \):
\[ a = \frac{3u^2}{2S} \tag{1} \]

### Step 3: Apply the second case (from speed \( 2u \) to final speed after covering additional distance \( S \))
Now we need to find the final speed after covering an additional distance \( S \). The total distance covered from the initial point is \( 2S \). We use the same equation of motion:
\[ v^2 = u^2 + 2as \]
In this case, the initial speed \( u = 2u \), the final speed \( v \) is what we want to find, and the distance \( s = 2S \):
\[ v^2 = (2u)^2 + 2a(2S) \]
Substituting \( a \) from equation (1):
\[ v^2 = 4u^2 + 2 \left(\frac{3u^2}{2S}\right)(2S) \]
This simplifies to:
\[ v^2 = 4u^2 + 3u^2 = 7u^2 \]
Taking the square root gives:
\[ v = \sqrt{7}u \]

### Final Answer
The speed of the body after covering an additional distance \( S \) is:
\[ \boxed{\sqrt{7}u} \]