The average velocity of a body moving with uniform acceleration after travelling a distance of `3.06 m` is `0.34 m s^(-1)`. If the change in velocity of the body is `0.18 ms^(-1)` during this time, its uniform acceleration is .

Options

A. `0.01 ms^(-2)`
B. `0.02 ms^(-2)`
C. `0.03 ms^(-2)`
D. `0.04 ms^(-2)`

Explanation

To find the uniform acceleration of the body, we can use the information provided in the question. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Distance (s) = 3.06 m
- Average velocity (v_avg) = 0.34 m/s
- Change in velocity (Δv) = 0.18 m/s

### Step 2: Calculate the time taken (t)
We can use the formula for average velocity:
\[ v_{avg} = \frac{s}{t} \]
Rearranging this formula to find time (t):
\[ t = \frac{s}{v_{avg}} = \frac{3.06 \, \text{m}}{0.34 \, \text{m/s}} \approx 9 \, \text{s} \]

### Step 3: Use the formula for acceleration
Acceleration (a) can be calculated using the formula:
\[ a = \frac{\Delta v}{t} \]
Where:
- Δv = change in velocity
- t = time taken

### Step 4: Substitute the values into the acceleration formula
Substituting the known values:
\[ a = \frac{0.18 \, \text{m/s}}{9 \, \text{s}} = 0.02 \, \text{m/s}^2 \]

### Final Answer:
The uniform acceleration of the body is \(0.02 \, \text{m/s}^2\).

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