A particle moving in a straight line covers half the distance with speed `v_(0)`. The other half of the distance is covered in two equal time intervals with speed `v_(1)` and `v_(2)`, respectively. The average speed of the particle during this motion is

Options

A. `(v_(0)(v_(1)+v_(2)))/(v_(0)+v_(1)+v_(2))`
B. `(2v_(0)(v_(1)+v_(2)))/(v_(0)+v_(1)+v_(2))`
C. `(2v_(0)(v_(1)+v_(2)))/(2v_(0)+v_(1)+v_(2))`
D. `(v_(0)(v_(1)+v_(2)))/(2v_(0)+v_(1)+v_(2))`

Explanation

To find the average speed of the particle during its motion, we can follow these steps:

### Step 1: Define the total distance and segments
Let the total distance covered by the particle be \( d \). According to the problem, the particle covers half of this distance, which is \( \frac{d}{2} \), at a speed of \( v_0 \). The other half of the distance, also \( \frac{d}{2} \), is covered in two equal time intervals with speeds \( v_1 \) and \( v_2 \).

### Step 2: Calculate the time taken for the first half
The time taken to cover the first half of the distance \( \frac{d}{2} \) at speed \( v_0 \) can be calculated using the formula:
\[ t_1 = \frac{\text{distance}}{\text{speed}} = \frac{\frac{d}{2}}{v_0} = \frac{d}{2v_0} \]

### Step 3: Define the time intervals for the second half
Let the time taken for each of the two equal time intervals be \( t \). Thus, the total time for the second half of the distance is \( 2t \).

### Step 4: Calculate the distance covered in the second half
In the second half of the distance, the particle covers \( \frac{d}{2} \) in two segments with speeds \( v_1 \) and \( v_2 \). The distances covered in each segment can be expressed as:
- Distance covered in the first interval: \( d_1 = v_1 t \)
- Distance covered in the second interval: \( d_2 = v_2 t \)

Since the total distance for the second half is \( \frac{d}{2} \), we have:
\[ d_1 + d_2 = \frac{d}{2} \]
This gives us:
\[ v_1 t + v_2 t = \frac{d}{2} \]
Factoring out \( t \):
\[ t(v_1 + v_2) = \frac{d}{2} \]
Thus, we can solve for \( t \):
\[ t = \frac{d}{2(v_1 + v_2)} \]

### Step 5: Calculate the total time taken
The total time \( T \) taken for the entire journey is the sum of the time taken for the first half and the time taken for the second half:
\[ T = t_1 + 2t = \frac{d}{2v_0} + 2\left(\frac{d}{2(v_1 + v_2)}\right) \]
This simplifies to:
\[ T = \frac{d}{2v_0} + \frac{d}{v_1 + v_2} \]

### Step 6: Calculate the average speed
The average speed \( v_{avg} \) is defined as the total distance divided by the total time:
\[ v_{avg} = \frac{\text{total distance}}{\text{total time}} = \frac{d}{T} \]
Substituting for \( T \):
\[ v_{avg} = \frac{d}{\frac{d}{2v_0} + \frac{d}{v_1 + v_2}} \]
Factoring out \( d \) from the denominator:
\[ v_{avg} = \frac{1}{\frac{1}{2v_0} + \frac{1}{v_1 + v_2}} \]
To combine the fractions, we find a common denominator:
\[ v_{avg} = \frac{1}{\frac{(v_1 + v_2) + 2v_0}{2v_0(v_1 + v_2)}} \]
This simplifies to:
\[ v_{avg} = \frac{2v_0(v_1 + v_2)}{(v_1 + v_2) + 2v_0} \]

### Final Answer
The average speed of the particle during its motion is:
\[ v_{avg} = \frac{2v_0(v_1 + v_2)}{(v_1 + v_2) + 2v_0} \]