In a typical lightning flash, the potential difference between discharge points is about `1.0 xx 10^(9) V` and the quantity of charge transferred I about 30 C. IF all the energy released could be used to accelerate a 1200 kg automobile from rest, what would be the final speed of the automobile ?

Options

A. `7100 ms^(-1)`
B. `3600 ms^(-1)`
C. `9500 ms^(-1)`
D. `6800 ms^(-1)`

Answer

To solve the problem step by step, we will use the principles of energy conservation and the relationship between electric potential energy and kinetic energy.

### Step 1: Calculate the energy released by the lightning flash
The energy (ΔU) released during the lightning flash can be calculated using the formula:

\[ \Delta U = Q \cdot \Delta V \]

Where:
- \( Q = 30 \, \text{C} \) (the quantity of charge transferred)
- \( \Delta V = 1.0 \times 10^9 \, \text{V} \) (the potential difference)

Substituting the values:

\[ \Delta U = 30 \, \text{C} \cdot 1.0 \times 10^9 \, \text{V} = 30 \times 10^9 \, \text{J} \]

### Step 2: Apply the conservation of energy
According to the conservation of energy, the energy transferred to the automobile will be converted into kinetic energy (KE). The kinetic energy of the automobile can be expressed as:

\[ KE = \frac{1}{2} m v^2 \]

Where:
- \( m = 1200 \, \text{kg} \) (mass of the automobile)
- \( v \) is the final speed we want to find.

Setting the energy from the lightning equal to the kinetic energy of the automobile:

\[ 30 \times 10^9 \, \text{J} = \frac{1}{2} \cdot 1200 \, \text{kg} \cdot v^2 \]

### Step 3: Solve for the final speed \( v \)
Rearranging the equation to solve for \( v^2 \):

\[ 30 \times 10^9 = 600 \cdot v^2 \]

Dividing both sides by 600:

\[ v^2 = \frac{30 \times 10^9}{600} \]

Calculating the right-hand side:

\[ v^2 = 50 \times 10^6 = 5 \times 10^7 \]

Now, taking the square root to find \( v \):

\[ v = \sqrt{5 \times 10^7} \approx 7071.06 \, \text{m/s} \]

### Step 4: Final answer
Rounding the value, we find:

\[ v \approx 7100 \, \text{m/s} \]

### Summary
The final speed of the automobile, if all the energy from the lightning flash is used to accelerate it, is approximately **7100 m/s**.

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