A body covers first one-third of the distance with a velocity 10 `ms^(-1)` in same direction, the second one-third with a velocity `20 ms^(-l)` and last one-third with a velocity of `30 ms^(-1)`. The average velocity of body is

Options

A. `17.8 ms^(-1)`
B. `16.4 ms^(-1)`
C. `18.3 ms^(-1)`
D. `20.2 ms^(-1)`

Answer

To find the average velocity of the body that covers different segments of distance with varying velocities, we can follow these steps:

### Step-by-Step Solution:

1. **Define the total distance**:
Let the total distance \( S \) be divided into three equal parts. Therefore, each part is \( \frac{S}{3} \).

2. **Calculate time for each segment**:
- For the first segment (velocity \( v_1 = 10 \, \text{m/s} \)):
\[ t_1 = \frac{\text{distance}}{\text{velocity}} = \frac{\frac{S}{3}}{10} = \frac{S}{30} \]
- For the second segment (velocity \( v_2 = 20 \, \text{m/s} \)):
\[ t_2 = \frac{\frac{S}{3}}{20} = \frac{S}{60} \]
- For the third segment (velocity \( v_3 = 30 \, \text{m/s} \)):
\[ t_3 = \frac{\frac{S}{3}}{30} = \frac{S}{90} \]

3. **Calculate total time taken**:
The total time \( T \) taken to cover the entire distance is the sum of the times for each segment:
\[ T = t_1 + t_2 + t_3 = \frac{S}{30} + \frac{S}{60} + \frac{S}{90} \]

4. **Find a common denominator**:
The least common multiple of 30, 60, and 90 is 180. Thus, we can rewrite each term:
\[ T = \frac{6S}{180} + \frac{3S}{180} + \frac{2S}{180} = \frac{(6 + 3 + 2)S}{180} = \frac{11S}{180} \]

5. **Calculate average velocity**:
The average velocity \( V_{avg} \) is defined as the total displacement divided by the total time taken:
\[ V_{avg} = \frac{S}{T} = \frac{S}{\frac{11S}{180}} = \frac{180}{11} \, \text{m/s} \]

6. **Final calculation**:
\[ V_{avg} \approx 16.36 \, \text{m/s} \]

### Conclusion:
The average velocity of the body is approximately \( 16.36 \, \text{m/s} \).

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