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 three equal distances with different velocities, we can follow these steps:

### Step 1: Define the total distance
Let the total distance covered by the body be \( D \). Since the body covers this distance in three equal parts, each part will be:
\[ d = \frac{D}{3} \]

### Step 2: Calculate the time taken for each segment
The time taken for each segment can be calculated using the formula:
\[ \text{Time} = \frac{\text{Distance}}{\text{Velocity}} \]

1. For the first segment (velocity = 10 m/s):
\[ T_1 = \frac{d}{10} = \frac{D/3}{10} = \frac{D}{30} \]

2. For the second segment (velocity = 20 m/s):
\[ T_2 = \frac{d}{20} = \frac{D/3}{20} = \frac{D}{60} \]

3. For the third segment (velocity = 30 m/s):
\[ T_3 = \frac{d}{30} = \frac{D/3}{30} = \frac{D}{90} \]

### Step 3: Calculate the total time taken
Now, we can find the total time taken \( T \) by summing up the times for each segment:
\[ T = T_1 + T_2 + T_3 = \frac{D}{30} + \frac{D}{60} + \frac{D}{90} \]

### Step 4: Find a common denominator and simplify
To add these fractions, we need a common denominator. The least common multiple of 30, 60, and 90 is 180. Thus, we can rewrite the times:
\[ T = \frac{6D}{180} + \frac{3D}{180} + \frac{2D}{180} = \frac{(6 + 3 + 2)D}{180} = \frac{11D}{180} \]

### Step 5: Calculate the average velocity
The average velocity \( V_{avg} \) is given by the formula:
\[ V_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{D}{T} \]
Substituting the total time we found:
\[ V_{avg} = \frac{D}{\frac{11D}{180}} = \frac{180}{11} \approx 16.36 \, \text{m/s} \]

### Conclusion
Thus, the average velocity of the body is approximately:
\[ V_{avg} \approx 16.36 \, \text{m/s} \]