A man who is running has half the kinetic energy of a boy of half his mass. The man speeds up by 1 `ms^(-1)` and then has the same kinetic energy as the boy. The original speeds of the man and the boy was:

Options

A. `2.4 m s^(-1), 1.2 m s^(-1)`
B. `1.2 m s^(-1), 4.4 m s^(-1)`
C. `2.4 m s^(-1), 4.8 m s^(-1)`
D. `4.8 m s^(-1), 2.4 m s^(-1)`

Explanation

To solve the problem step by step, let's denote the mass of the man as \( m \) and the mass of the boy as \( \frac{m}{2} \). Let the velocity of the man be \( V_m \) and the velocity of the boy be \( V_b \).

### Step 1: Write the kinetic energy equations
The kinetic energy (KE) of the man is given by:
\[ KE_m = \frac{1}{2} m V_m^2 \]
The kinetic energy of the boy is:
\[ KE_b = \frac{1}{2} \left(\frac{m}{2}\right) V_b^2 = \frac{1}{4} m V_b^2 \]

### Step 2: Set up the relationship between their kinetic energies
According to the problem, the man has half the kinetic energy of the boy:
\[ KE_m = \frac{1}{2} KE_b \]
Substituting the expressions for kinetic energy, we have:
\[ \frac{1}{2} m V_m^2 = \frac{1}{2} \left(\frac{1}{4} m V_b^2\right) \]
This simplifies to:
\[ m V_m^2 = \frac{1}{4} m V_b^2 \]
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ V_m^2 = \frac{1}{4} V_b^2 \]

### Step 3: Relate the velocities
Taking the square root of both sides gives:
\[ V_m = \frac{1}{2} V_b \quad \text{(Equation 1)} \]

### Step 4: Consider the change in the man's speed
When the man speeds up by \( 1 \, \text{m/s} \), his new velocity becomes \( V_m + 1 \). The new kinetic energy of the man is:
\[ KE_m' = \frac{1}{2} m (V_m + 1)^2 \]
The kinetic energy of the boy remains:
\[ KE_b = \frac{1}{4} m V_b^2 \]

### Step 5: Set the new kinetic energy of the man equal to that of the boy
According to the problem, after speeding up, the man has the same kinetic energy as the boy:
\[ \frac{1}{2} m (V_m + 1)^2 = \frac{1}{4} m V_b^2 \]
Dividing both sides by \( \frac{1}{2} m \):
\[ (V_m + 1)^2 = \frac{1}{2} V_b^2 \]

### Step 6: Substitute \( V_b \) from Equation 1
From Equation 1, we know \( V_b = 2 V_m \). Substituting this into the equation gives:
\[ (V_m + 1)^2 = \frac{1}{2} (2 V_m)^2 \]
This simplifies to:
\[ (V_m + 1)^2 = 2 V_m^2 \]

### Step 7: Expand and simplify
Expanding the left side:
\[ V_m^2 + 2V_m + 1 = 2 V_m^2 \]
Rearranging gives:
\[ 0 = 2 V_m^2 - V_m^2 - 2V_m - 1 \]
\[ 0 = V_m^2 - 2V_m - 1 \]

### Step 8: Solve the quadratic equation
Using the quadratic formula \( V_m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -2, c = -1 \):
\[ V_m = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
\[ V_m = \frac{2 \pm \sqrt{4 + 4}}{2} \]
\[ V_m = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2} \]
Taking the positive root:
\[ V_m = 1 + \sqrt{2} \]

### Step 9: Calculate \( V_b \)
Using Equation 1:
\[ V_b = 2V_m = 2(1 + \sqrt{2}) = 2 + 2\sqrt{2} \]

### Final Results
Calculating the approximate values:
- \( V_m \approx 2.414 \, \text{m/s} \)
- \( V_b \approx 4.828 \, \text{m/s} \)

### Conclusion
The original speeds of the man and the boy are approximately:
- \( V_m \approx 2.4 \, \text{m/s} \)
- \( V_b \approx 4.8 \, \text{m/s} \)