If the speed of a vehicle is increased by `1 ms^(-1)`, its kinetic energy is doubled , then original speed of the vehicle is

Options

A. `(sqrt2+1)ms^(-1)`
B. `2(sqrt2-1)ms^(-1)`
C. `2(sqrt2+1)ms^(-1)`
D. `sqrt2(sqrt2+1)ms^(-1)`

Explanation

To solve the problem, we will use the formula for kinetic energy and set up an equation based on the information provided.

### Step-by-Step Solution:

1. **Define the Variables:**
Let the original speed of the vehicle be \( V \) m/s. The kinetic energy (KE) of the vehicle at this speed is given by the formula:
\[ KE_{\text{initial}} = \frac{1}{2} m V^2 \]
where \( m \) is the mass of the vehicle.

2. **Determine the Final Speed:**
If the speed of the vehicle is increased by \( 1 \) m/s, the new speed becomes:
\[ V_{\text{final}} = V + 1 \text{ m/s} \]

3. **Calculate the Final Kinetic Energy:**
The kinetic energy at the new speed is:
\[ KE_{\text{final}} = \frac{1}{2} m (V + 1)^2 \]

4. **Set Up the Equation:**
According to the problem, the final kinetic energy is double the initial kinetic energy:
\[ KE_{\text{final}} = 2 \cdot KE_{\text{initial}} \]
Substituting the expressions for kinetic energy, we have:
\[ \frac{1}{2} m (V + 1)^2 = 2 \cdot \frac{1}{2} m V^2 \]

5. **Simplify the Equation:**
The mass \( m \) and \( \frac{1}{2} \) can be canceled from both sides:
\[ (V + 1)^2 = 2V^2 \]

6. **Expand and Rearrange:**
Expanding the left side gives:
\[ V^2 + 2V + 1 = 2V^2 \]
Rearranging this equation results in:
\[ 0 = 2V^2 - V^2 - 2V - 1 \]
which simplifies to:
\[ V^2 - 2V - 1 = 0 \]

7. **Solve the Quadratic Equation:**
We can use the quadratic formula \( V = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -2, c = -1 \):
\[ V = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
\[ V = \frac{2 \pm \sqrt{4 + 4}}{2} \]
\[ V = \frac{2 \pm \sqrt{8}}{2} \]
\[ V = \frac{2 \pm 2\sqrt{2}}{2} \]
\[ V = 1 \pm \sqrt{2} \]

8. **Select the Positive Root:**
Since speed cannot be negative, we take:
\[ V = 1 + \sqrt{2} \text{ m/s} \]

### Final Answer:
The original speed of the vehicle is \( 1 + \sqrt{2} \) m/s.

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