In a particle accelerator , an electron enters a region in which it accelerates uniformly in a straight line from a speed of `4.00xx10^(5)m//s` to a speed of `6.00xx10^(7)m//s` in a distance of 3.00 cm. For what time interval does the electron accelerate ?

Answer

To solve the problem of an electron accelerating uniformly in a straight line from an initial speed of \(4.00 \times 10^5 \, \text{m/s}\) to a final speed of \(6.00 \times 10^7 \, \text{m/s}\) over a distance of \(3.00 \, \text{cm}\), we can follow these steps:

### Step 1: Convert Distance to Meters
First, we need to convert the distance from centimeters to meters since the speeds are given in meters per second.

\[ 3.00 \, \text{cm} = 0.03 \, \text{m} \]

### Step 2: Use the Kinematic Equation
We can use the kinematic equation that relates initial velocity (\(u\)), final velocity (\(v\)), acceleration (\(a\)), and distance (\(s\)):

\[ v^2 = u^2 + 2as \]

### Step 3: Substitute Known Values
Substituting the known values into the equation:

- \(u = 4.00 \times 10^5 \, \text{m/s}\)
- \(v = 6.00 \times 10^7 \, \text{m/s}\)
- \(s = 0.03 \, \text{m}\)

We have:

\[ (6.00 \times 10^7)^2 = (4.00 \times 10^5)^2 + 2a(0.03) \]

### Step 4: Calculate \(v^2\) and \(u^2\)
Calculating \(v^2\) and \(u^2\):

\[ (6.00 \times 10^7)^2 = 36.00 \times 10^{14} \, \text{m}^2/\text{s}^2 \]
\[ (4.00 \times 10^5)^2 = 16.00 \times 10^{10} \, \text{m}^2/\text{s}^2 \]

### Step 5: Substitute and Rearrange for Acceleration
Now substitute these values back into the equation:

\[ 36.00 \times 10^{14} = 16.00 \times 10^{10} + 2a(0.03) \]

Rearranging gives:

\[ 2a(0.03) = 36.00 \times 10^{14} - 16.00 \times 10^{10} \]

### Step 6: Simplify the Right Side
To simplify, we need to express both terms with the same power of ten. We can express \(36.00 \times 10^{14}\) as \(3600.00 \times 10^{12}\) and \(16.00 \times 10^{10}\) as \(0.16 \times 10^{12}\):

\[ 2a(0.03) = (3600.00 - 0.16) \times 10^{12} \]
\[ 2a(0.03) = 3599.84 \times 10^{12} \]

### Step 7: Solve for Acceleration \(a\)
Now we can solve for \(a\):

\[ a = \frac{3599.84 \times 10^{12}}{2 \times 0.03} \]
\[ a = \frac{3599.84 \times 10^{12}}{0.06} \]
\[ a = 59997.33 \times 10^{12} \, \text{m/s}^2 \]

### Step 8: Use the Second Kinematic Equation to Find Time
Now we can use the equation:

\[ v = u + at \]

Rearranging for \(t\):

\[ t = \frac{v - u}{a} \]

Substituting the values:

\[ t = \frac{6.00 \times 10^7 - 4.00 \times 10^5}{59997.33 \times 10^{12}} \]

### Step 9: Calculate the Time Interval
Calculating the numerator:

\[ 6.00 \times 10^7 - 4.00 \times 10^5 = 59996.00 \times 10^5 \]

Now substituting back into the equation:

\[ t = \frac{59996.00 \times 10^5}{59997.33 \times 10^{12}} \]

### Step 10: Final Calculation
This simplifies to:

\[ t \approx 0.000999 \, \text{s} \text{ or } 9.99 \times 10^{-4} \, \text{s} \]

### Final Answer
The time interval during which the electron accelerates is approximately:

\[ t \approx 9.99 \times 10^{-4} \, \text{s} \]

---