Two particles move along x-axis in the same direction with uniform velocities `8 m//s` and `4 m//s`. Initially the first particle is 21 m to the left of the origin and the second one is 7 m to the right of the origin. The two particles meet from the origin at a distance of

Options

A. 35 m
B. 32 m
C. 28 m
D. 56 m

Explanation

To solve the problem, we will follow these steps:

### Step 1: Define the initial positions of the particles
- Particle 1 (P1) is initially at \( x_1 = -21 \, \text{m} \) (21 m to the left of the origin).
- Particle 2 (P2) is initially at \( x_2 = 7 \, \text{m} \) (7 m to the right of the origin).

### Step 2: Define the velocities of the particles
- Velocity of Particle 1, \( v_1 = 8 \, \text{m/s} \)
- Velocity of Particle 2, \( v_2 = 4 \, \text{m/s} \)

### Step 3: Set up the equations for the positions of the particles over time
Let \( t \) be the time in seconds after they start moving. The positions of the particles at time \( t \) can be expressed as:
- Position of Particle 1:
\[ x_1(t) = -21 + 8t \]
- Position of Particle 2:
\[ x_2(t) = 7 + 4t \]

### Step 4: Set the positions equal to find when they meet
To find the time \( t \) when the two particles meet, we set their position equations equal to each other:
\[ -21 + 8t = 7 + 4t \]

### Step 5: Solve for \( t \)
Rearranging the equation gives:
\[ 8t - 4t = 7 + 21 \]
\[ 4t = 28 \]
\[ t = 7 \, \text{s} \]

### Step 6: Find the meeting point
Now, we can substitute \( t = 7 \, \text{s} \) back into either position equation to find the distance from the origin where they meet. Using Particle 2's position:
\[ x_2(7) = 7 + 4(7) = 7 + 28 = 35 \, \text{m} \]

### Final Answer
The two particles meet at a distance of **35 meters from the origin**.

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