A particle moving in a straight line covers first half of the distance into two equal time intervals with speed of 2 m/s and 4 m/s respectively, and second half with the speed of 5 m/s. The average speed of the particle is

Options

A. 4.75 m/s
B. 4.25 m/s
C. 4.00 m/s
D. 3.75 m/s

Answer

To find the average speed of the particle, we need to follow these steps:

### Step 1: Define the total distance
Let the total distance covered by the particle be \( D \). The first half of the distance is \( \frac{D}{2} \) and the second half is also \( \frac{D}{2} \).

### Step 2: Calculate time taken for the first half of the distance
The first half of the distance is covered in two equal time intervals, which we will denote as \( t \).

1. In the first time interval \( t \), the particle travels a distance \( d_1 \) with a speed of \( 2 \, \text{m/s} \):
\[ d_1 = v_1 \cdot t = 2 \cdot t = 2t \]

2. In the second time interval \( t \), the particle travels a distance \( d_2 \) with a speed of \( 4 \, \text{m/s} \):
\[ d_2 = v_2 \cdot t = 4 \cdot t = 4t \]

3. The total distance for the first half is:
\[ d_1 + d_2 = 2t + 4t = 6t \]

### Step 3: Relate the first half distance to \( D \)
Since the first half of the distance is \( \frac{D}{2} \):
\[ 6t = \frac{D}{2} \implies D = 12t \]

### Step 4: Calculate the time taken for the second half of the distance
The second half of the distance \( \frac{D}{2} \) is covered with a speed of \( 5 \, \text{m/s} \):
\[ \text{Time for second half} = \frac{\text{Distance}}{\text{Speed}} = \frac{\frac{D}{2}}{5} = \frac{12t/2}{5} = \frac{6t}{5} \]

### Step 5: Calculate the total time taken
The total time taken \( T \) is the sum of the time taken for both halves:
\[ T = \text{Time for first half} + \text{Time for second half} = 2t + \frac{6t}{5} \]

To combine these, convert \( 2t \) into a fraction with a common denominator:
\[ 2t = \frac{10t}{5} \]
So,
\[ T = \frac{10t}{5} + \frac{6t}{5} = \frac{16t}{5} \]

### Step 6: Calculate the average speed
The average speed \( V_{avg} \) is given by the total distance divided by the total time:
\[ V_{avg} = \frac{D}{T} = \frac{12t}{\frac{16t}{5}} = 12t \cdot \frac{5}{16t} = \frac{60}{16} = 3.75 \, \text{m/s} \]

### Final Answer
The average speed of the particle is \( 3.75 \, \text{m/s} \).

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