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 need to find the original speed of the vehicle given that increasing its speed by \(1 \, \text{m/s}\) doubles its kinetic energy. Let's break down the solution step by step.

### Step 1: Define the initial kinetic energy
The initial kinetic energy (\(KE\)) of the vehicle when its speed is \(v\) is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
where \(m\) is the mass of the vehicle.

### Step 2: Define the new speed and kinetic energy
When the speed is increased by \(1 \, \text{m/s}\), the new speed becomes \(v + 1\). The new kinetic energy (\(KE'\)) is:
\[ KE' = \frac{1}{2} m(v + 1)^2 \]

### Step 3: Set up the equation based on the problem statement
According to the problem, the new kinetic energy is double the initial kinetic energy:
\[ KE' = 2 \times KE \]
Substituting the expressions for kinetic energy, we get:
\[ \frac{1}{2} m(v + 1)^2 = 2 \times \frac{1}{2} mv^2 \]

### Step 4: Simplify the equation
We can cancel \(\frac{1}{2} m\) from both sides (assuming \(m \neq 0\)):
\[ (v + 1)^2 = 2v^2 \]

### Step 5: Expand and rearrange the equation
Expanding the left side:
\[ v^2 + 2v + 1 = 2v^2 \]
Rearranging gives:
\[ 0 = 2v^2 - v^2 - 2v - 1 \]
which simplifies to:
\[ v^2 - 2v - 1 = 0 \]

### Step 6: Solve the quadratic equation
We can use the quadratic formula \(v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = -2\), and \(c = -1\):
\[ v = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
\[ v = \frac{2 \pm \sqrt{4 + 4}}{2} \]
\[ v = \frac{2 \pm \sqrt{8}}{2} \]
\[ v = \frac{2 \pm 2\sqrt{2}}{2} \]
\[ v = 1 \pm \sqrt{2} \]

### Step 7: Determine the valid solution
Since speed cannot be negative, we take the positive solution:
\[ v = 1 + \sqrt{2} \]

### Final Answer
The original speed of the vehicle is:
\[ \boxed{1 + \sqrt{2} \, \text{m/s}} \]