The K.E. of a person is just half of K.E. of a boy whose mass is just half of that person. If person increases its speed by 1 m/s, then its K.E. equals to that of boy then initial speed of person was -

Options

A. `(sqrt(2)+1)`m/s
B. `(2+sqrt(2))` m/s
C. `2(sqrt(2)+2)` m/s
D. None

Explanation

To solve the problem step by step, we need to analyze the relationship between the kinetic energies of the person and the boy, as well as their masses and speeds.

### Step 1: Define the Variables
Let:
- \( m_p \) = mass of the person
- \( m_b \) = mass of the boy = \( \frac{1}{2} m_p \) (since the boy's mass is half of the person's mass)
- \( v_p \) = initial speed of the person
- \( v_b \) = speed of the boy (unknown for now)

### Step 2: Write the Kinetic Energy Equations
The kinetic energy (K.E.) of the person is given by:
\[ K.E._p = \frac{1}{2} m_p v_p^2 \]
The kinetic energy of the boy is:
\[ K.E._b = \frac{1}{2} m_b v_b^2 = \frac{1}{2} \left(\frac{1}{2} m_p\right) v_b^2 = \frac{1}{4} m_p v_b^2 \]
According to the problem, the kinetic energy of the person is half that of the boy:
\[ K.E._p = \frac{1}{2} K.E._b \]
Substituting the expressions for kinetic energy:
\[ \frac{1}{2} m_p v_p^2 = \frac{1}{2} \left(\frac{1}{4} m_p v_b^2\right) \]
This simplifies to:
\[ m_p v_p^2 = \frac{1}{4} m_p v_b^2 \]
Cancelling \( m_p \) (assuming \( m_p \neq 0 \)):
\[ v_p^2 = \frac{1}{4} v_b^2 \]
Taking the square root:
\[ v_p = \frac{1}{2} v_b \]

### Step 3: Kinetic Energy After Speed Increase
When the person increases their speed by 1 m/s, their new speed becomes:
\[ v_p + 1 \]
The new kinetic energy of the person is:
\[ K.E._p' = \frac{1}{2} m_p (v_p + 1)^2 \]
According to the problem, this new kinetic energy equals the kinetic energy of the boy:
\[ K.E._p' = K.E._b \]
Substituting the expressions:
\[ \frac{1}{2} m_p (v_p + 1)^2 = \frac{1}{4} m_p v_b^2 \]
Cancelling \( m_p \) (again assuming \( m_p \neq 0 \)):
\[ (v_p + 1)^2 = \frac{1}{2} v_b^2 \]

### Step 4: Substitute \( v_b \)
From the earlier step, we have \( v_b = 2v_p \). Substituting this into the equation:
\[ (v_p + 1)^2 = \frac{1}{2} (2v_p)^2 \]
This simplifies to:
\[ (v_p + 1)^2 = 2v_p^2 \]

### Step 5: Expand and Rearrange
Expanding the left side:
\[ v_p^2 + 2v_p + 1 = 2v_p^2 \]
Rearranging gives:
\[ 0 = 2v_p^2 - v_p^2 - 2v_p - 1 \]
\[ 0 = v_p^2 - 2v_p - 1 \]

### Step 6: Solve the Quadratic Equation
Using the quadratic formula \( v_p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -2, c = -1 \):
\[ v_p = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
\[ v_p = \frac{2 \pm \sqrt{4 + 4}}{2} \]
\[ v_p = \frac{2 \pm \sqrt{8}}{2} \]
\[ v_p = \frac{2 \pm 2\sqrt{2}}{2} \]
\[ v_p = 1 \pm \sqrt{2} \]

### Step 7: Determine the Positive Solution
Since speed cannot be negative, we take:
\[ v_p = 1 + \sqrt{2} \]

### Final Answer
The initial speed of the person is:
\[
\boxed{1 + \sqrt{2}} \text{ m/s}