If a man increase his speed by 2m/s, his K.E. is doubled, the original speed of the man is

Options

A. `(1+2sqrt(2))m//s`
B. `4m//s`
C. `(2+2sqrt(2))m//s`
D. `(2+sqrt(2))m//s`

Answer

To solve the problem, we need to find the original speed of a man given that increasing his speed by 2 m/s doubles his kinetic energy.

### Step-by-Step Solution:

1. **Define Variables**:
Let the original speed of the man be \( v \) m/s.

2. **Write the Expression for Original Kinetic Energy**:
The original kinetic energy (K.E.) of the man can be expressed as:
\[ K.E._1 = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the man.

3. **Express the New Speed**:
When the man increases his speed by 2 m/s, his new speed becomes:
\[ v + 2 \text{ m/s} \]

4. **Write the Expression for New Kinetic Energy**:
The new kinetic energy after the speed increase is:
\[ K.E._2 = \frac{1}{2} m(v + 2)^2 \]

5. **Set Up the Equation for Kinetic Energy**:
According to the problem, the new kinetic energy is double the original kinetic energy:
\[ K.E._2 = 2 \times K.E._1 \]
Substituting the expressions we have:
\[ \frac{1}{2} m(v + 2)^2 = 2 \times \frac{1}{2} mv^2 \]

6. **Simplify the Equation**:
The \( \frac{1}{2} m \) on both sides cancels out, leading to:
\[ (v + 2)^2 = 2v^2 \]

7. **Expand the Left Side**:
Expanding \( (v + 2)^2 \):
\[ v^2 + 4v + 4 = 2v^2 \]

8. **Rearrange the Equation**:
Rearranging gives:
\[ 0 = 2v^2 - v^2 - 4v - 4 \]
Simplifying further:
\[ v^2 - 4v - 4 = 0 \]

9. **Use the Quadratic Formula**:
To solve for \( v \), we can use the quadratic formula:
\[ v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -4 \), and \( c = -4 \):
\[ v = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} \]
\[ v = \frac{4 \pm \sqrt{16 + 16}}{2} \]
\[ v = \frac{4 \pm \sqrt{32}}{2} \]
\[ v = \frac{4 \pm 4\sqrt{2}}{2} \]
\[ v = 2 \pm 2\sqrt{2} \]

10. **Select the Positive Solution**:
Since speed cannot be negative, we take:
\[ v = 2 + 2\sqrt{2} \]

### Final Answer:
The original speed of the man is \( 2 + 2\sqrt{2} \) m/s.