A proton moving along the `x` axis has an initial velocity of `4.0xx10^(6)m//s` and a constant acceleration of `6.0xx10^(12)m//s^(2)`. What is the velocity of the proton after it has traveled a distance of `80 cm` ?

Options

A. `5.1xx10^(6)m//s`
B. `6.3 xx 10^(6)m//s`
C. `4.8 xx10^(6)m//s`
D. `3.9 xx 10^(6) m//s`

Answer

To solve the problem of finding the final velocity of a proton moving along the x-axis with a given initial velocity, constant acceleration, and distance traveled, we can use the equations of motion. Here’s a step-by-step solution:

### Step 1: Identify the known values
- Initial velocity (u) = \(4.0 \times 10^6 \, \text{m/s}\)
- Constant acceleration (a) = \(6.0 \times 10^{12} \, \text{m/s}^2\)
- Distance traveled (s) = \(80 \, \text{cm} = 80 \times 10^{-2} \, \text{m} = 0.8 \, \text{m}\)

### Step 2: Use the equation of motion
We will use the third equation of motion, which relates initial velocity, final velocity, acceleration, and distance:
\[ V^2 = U^2 + 2AS \]
Where:
- \(V\) = final velocity
- \(U\) = initial velocity
- \(A\) = acceleration
- \(S\) = distance traveled

### Step 3: Substitute the known values into the equation
Substituting the values we have:
\[ V^2 = (4.0 \times 10^6)^2 + 2 \times (6.0 \times 10^{12}) \times (0.8) \]

### Step 4: Calculate \(U^2\)
Calculating \(U^2\):
\[ (4.0 \times 10^6)^2 = 16.0 \times 10^{12} \, \text{m}^2/\text{s}^2 \]

### Step 5: Calculate \(2AS\)
Calculating \(2AS\):
\[ 2 \times (6.0 \times 10^{12}) \times (0.8) = 9.6 \times 10^{12} \, \text{m}^2/\text{s}^2 \]

### Step 6: Combine the results
Now, we can combine the results:
\[ V^2 = 16.0 \times 10^{12} + 9.6 \times 10^{12} = 25.6 \times 10^{12} \, \text{m}^2/\text{s}^2 \]

### Step 7: Calculate \(V\)
Taking the square root to find \(V\):
\[ V = \sqrt{25.6 \times 10^{12}} = 5.06 \times 10^6 \, \text{m/s} \]

### Step 8: Round the result
Rounding the result gives:
\[ V \approx 5.1 \times 10^6 \, \text{m/s} \]

### Final Answer
The final velocity of the proton after it has traveled a distance of 80 cm is approximately \(5.1 \times 10^6 \, \text{m/s}\).

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