If a car covers `(2)/(3)` of the total distance with speed `nu_(1)` and `(3)/(5)` distance with speed v, then average speed is

Options

A. `(1)/(2)sqrt(nu_(1)nu_(2))`
B. `(nu_(1)+nu_(2))/(2)`
C. `(2nu_(1)nu_(2))/(nu_(1)+nu_(2))`
D. `(5nu_(1)nu_(2))/(3nu_(1)+2nu_(2))`

Explanation

To find the average speed of the car that covers different fractions of the total distance with different speeds, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Total Distance**:
Let the total distance be \( S \).

2. **Calculate Distances Covered**:
The car covers \( \frac{2}{5}S \) of the total distance with speed \( v_1 \) and \( \frac{3}{5}S \) of the total distance with speed \( v_2 \).

3. **Calculate Time Taken for Each Segment**:
- For the first segment (distance \( \frac{2}{5}S \) with speed \( v_1 \)):
\[ t_1 = \frac{\text{Distance}}{\text{Speed}} = \frac{\frac{2}{5}S}{v_1} = \frac{2S}{5v_1} \]
- For the second segment (distance \( \frac{3}{5}S \) with speed \( v_2 \)):
\[ t_2 = \frac{\frac{3}{5}S}{v_2} = \frac{3S}{5v_2} \]

4. **Calculate Total Time Taken**:
The total time \( T \) taken for the journey is the sum of the times for both segments:
\[ T = t_1 + t_2 = \frac{2S}{5v_1} + \frac{3S}{5v_2} \]

5. **Factor Out the Common Terms**:
We can factor out \( S \) from the total time:
\[ T = S \left( \frac{2}{5v_1} + \frac{3}{5v_2} \right) = S \cdot \frac{1}{5} \left( \frac{2}{v_1} + \frac{3}{v_2} \right) \]

6. **Calculate Average Speed**:
The average speed \( V_{avg} \) is defined as the total distance divided by the total time:
\[ V_{avg} = \frac{S}{T} = \frac{S}{S \cdot \frac{1}{5} \left( \frac{2}{v_1} + \frac{3}{v_2} \right)} = \frac{5}{\frac{2}{v_1} + \frac{3}{v_2}} \]

7. **Simplify the Expression**:
To simplify further, we can find a common denominator:
\[ V_{avg} = \frac{5v_1v_2}{3v_1 + 2v_2} \]

### Final Answer:
The average speed of the car is:
\[ V_{avg} = \frac{5v_1v_2}{3v_1 + 2v_2} \]