A muon ( an elementary particle ) enters a region with a speed of `6.00xx10^(6)` m/s and then is slowed at the rate of `1.25xx10^(14)m//s^(2)`. (a) How far does the muon take to stop ? (b) Graph `x` versus t and v versus t for the muon .

Answer

To solve the problem step by step, we will break it down into two parts as requested:

### Part (a): How far does the muon take to stop?

1. **Identify Given Values:**
- Initial velocity (u) = \(6.00 \times 10^6 \, \text{m/s}\)
- Final velocity (v) = \(0 \, \text{m/s}\) (since the muon stops)
- Acceleration (a) = \(-1.25 \times 10^{14} \, \text{m/s}^2\) (negative because it is deceleration)

2. **Use the Equation of Motion:**
We can use the equation of motion:
\[ v^2 = u^2 + 2as \]
where \(s\) is the distance traveled before stopping.

3. **Substitute the Known Values:**
\[ 0 = (6.00 \times 10^6)^2 + 2 \times (-1.25 \times 10^{14}) \times s \]

4. **Rearranging the Equation:**
\[ 0 = 36 \times 10^{12} - 2.5 \times 10^{14} s \]
\[ 2.5 \times 10^{14} s = 36 \times 10^{12} \]

5. **Solve for \(s\):**
\[ s = \frac{36 \times 10^{12}}{2.5 \times 10^{14}} = \frac{36}{2.5} \times 10^{-2} = 14.4 \times 10^{-2} = 0.144 \, \text{m} \]

### Answer for Part (a):
The muon takes **0.144 meters** to stop.

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### Part (b): Graph \(x\) versus \(t\) and \(v\) versus \(t\) for the muon.

1. **Velocity vs Time Graph:**
- We know that the initial velocity \(u = 6.00 \times 10^6 \, \text{m/s}\) and the final velocity \(v = 0 \, \text{m/s}\).
- The acceleration is negative, which indicates that the velocity decreases linearly over time.
- Using the equation \(v = u + at\):
\[ v = 6.00 \times 10^6 - (1.25 \times 10^{14})t \]
- When \(v = 0\):
\[ 0 = 6.00 \times 10^6 - (1.25 \times 10^{14})t \]
\[ t = \frac{6.00 \times 10^6}{1.25 \times 10^{14}} = 4.8 \times 10^{-8} \, \text{s} \]

2. **Plotting the Graph:**
- The graph of \(v\) versus \(t\) will be a straight line starting from \(6.00 \times 10^6 \, \text{m/s}\) at \(t = 0\) and reaching \(0 \, \text{m/s}\) at \(t = 4.8 \times 10^{-8} \, \text{s}\).
- The slope of the line will be negative, indicating deceleration.

3. **Position vs Time Graph:**
- To find the position as a function of time, we can use the equation:
\[ x = ut + \frac{1}{2}at^2 \]
- Substituting \(u\) and \(a\):
\[ x = (6.00 \times 10^6)t - \frac{1}{2}(1.25 \times 10^{14})t^2 \]
- The graph of \(x\) versus \(t\) will be a downward-opening parabola since the acceleration is negative.

### Summary of Graphs:
- **Velocity vs Time (v-t)**: A straight line decreasing from \(6.00 \times 10^6 \, \text{m/s}\) to \(0 \, \text{m/s}\).
- **Position vs Time (x-t)**: A downward-opening parabola starting from the origin.

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