Two particles are intially at a separation of d to each other. They move with constant velocirties with `5m//s` and `3m//s` respectively in horizontal plane. Their velocities makes an angle `60^(@)` with each other. If after 2 seconds, they collide. Then value of d is:

Options

A. `2sqrt(19)m`
B. `8.7m`
C. `2m`
D. `32m`

Explanation

To solve the problem step by step, we will follow the given information and apply the concepts of relative velocity and displacement.

### Step 1: Understand the Problem
We have two particles A and B moving with constant velocities of 5 m/s and 3 m/s, respectively, at an angle of 60° to each other. They collide after 2 seconds. We need to find the initial separation distance \( d \) between them.

### Step 2: Determine the Velocities
Let:
- \( V_A = 5 \, \text{m/s} \) (velocity of particle A)
- \( V_B = 3 \, \text{m/s} \) (velocity of particle B)
- The angle between their velocities \( \theta = 60^\circ \)

### Step 3: Calculate the Relative Velocity
The relative velocity \( V_{AB} \) of particle A with respect to particle B can be calculated using the formula:
\[ V_{AB} = \sqrt{V_A^2 + V_B^2 - 2 V_A V_B \cos(\theta)} \]

Substituting the values:
\[ V_{AB} = \sqrt{(5)^2 + (3)^2 - 2 \cdot 5 \cdot 3 \cdot \cos(60^\circ)} \]

Since \( \cos(60^\circ) = \frac{1}{2} \):
\[ V_{AB} = \sqrt{25 + 9 - 2 \cdot 5 \cdot 3 \cdot \frac{1}{2}} \]
\[ = \sqrt{25 + 9 - 15} \]
\[ = \sqrt{19} \, \text{m/s} \]

### Step 4: Calculate the Displacement Before Collision
The displacement \( S_{AB} \) between the two particles before they collide can be calculated using the formula:
\[ S_{AB} = V_{AB} \times t \]
where \( t = 2 \, \text{s} \).

Substituting the values:
\[ S_{AB} = \sqrt{19} \times 2 \]
\[ = 2\sqrt{19} \, \text{meters} \]

### Step 5: Calculate the Numerical Value
Now, we calculate \( 2\sqrt{19} \):
\[ \sqrt{19} \approx 4.3589 \, \text{(approximately)} \]
Thus,
\[ S_{AB} \approx 2 \times 4.3589 \approx 8.7178 \, \text{meters} \]

### Step 6: Conclusion
The initial separation distance \( d \) between the two particles is approximately:
\[ d \approx 8.7 \, \text{meters} \]

### Final Answer
The value of \( d \) is approximately **8.7 meters**.

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