A body travels uniformly a distance of `(10.2pm0.4)` m in a time interval `(6.0pm0.2)s`. The speed of the particle is best expressed as (in `m//s^(-1)`).

Options

A. `1.7pm0.12`
B. `1.7pm0.01`
C. `1.7pm0.07`
D. `1.7pm0.00`

Answer

To solve the problem of calculating the speed of a body that travels a distance of \(10.2 \pm 0.4\) m in a time interval of \(6.0 \pm 0.2\) s, we will follow these steps:

### Step 1: Calculate the Magnitude of Speed
The formula for speed \(v\) is given by:

\[ v = \frac{\text{distance}}{\text{time}} \]

Substituting the given values:

\[ v = \frac{10.2 \, \text{m}}{6.0 \, \text{s}} = 1.7 \, \text{m/s} \]

### Step 2: Calculate the Relative Errors
Next, we need to calculate the relative errors in distance and time.

1. **Relative error in distance**:
\[ \text{Relative error in distance} = \frac{\Delta s}{s} = \frac{0.4}{10.2} \]

2. **Relative error in time**:
\[ \text{Relative error in time} = \frac{\Delta t}{t} = \frac{0.2}{6.0} \]

### Step 3: Add the Relative Errors
Since speed is calculated by dividing distance by time, the relative errors will add up:

\[ \frac{\Delta v}{v} = \frac{\Delta s}{s} + \frac{\Delta t}{t} \]

### Step 4: Substitute the Values
Substituting the calculated relative errors:

1. Calculate relative error in distance:
\[ \frac{0.4}{10.2} \approx 0.0392 \]

2. Calculate relative error in time:
\[ \frac{0.2}{6.0} \approx 0.0333 \]

Adding these together:

\[ \frac{\Delta v}{v} \approx 0.0392 + 0.0333 \approx 0.0725 \]

### Step 5: Calculate the Absolute Error in Speed
Now, we can find the absolute error in speed:

\[ \Delta v = v \times \frac{\Delta v}{v} = 1.7 \times 0.0725 \approx 0.12325 \]

Rounding this to two decimal places gives us:

\[ \Delta v \approx 0.12 \]

### Step 6: Final Result
Thus, the speed of the particle can be expressed as:

\[ v = 1.7 \pm 0.12 \, \text{m/s} \]

### Summary
The final answer is:

\[ \text{Speed} = 1.7 \pm 0.12 \, \text{m/s} \]

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