A particle `A` moves with velocity `(2hati-3hatj)m//s` from a point `(4,5m)m`. At the same instant a particle `B`, moving in the same plane with velocity` (4hati+hatj)m//s` passes through a point `C(0,-3)m`. Find the `x`-coordinate (in `m`) of the point where the particles collide.

Answer

To find the x-coordinate of the point where particles A and B collide, we can follow these steps:

### Step 1: Write down the given information
- Particle A moves with a velocity of \( \vec{v_A} = (2 \hat{i} - 3 \hat{j}) \, \text{m/s} \) from the point \( (4, 5) \, \text{m} \).
- Particle B moves with a velocity of \( \vec{v_B} = (4 \hat{i} + \hat{j}) \, \text{m/s} \) from the point \( C(0, -3) \, \text{m} \).

### Step 2: Find the relative velocity of A with respect to B
The relative velocity of A with respect to B is given by:
\[ \vec{v_{AB}} = \vec{v_A} - \vec{v_B} \]
Substituting the values:
\[ \vec{v_{AB}} = (2 \hat{i} - 3 \hat{j}) - (4 \hat{i} + \hat{j}) = (2 - 4) \hat{i} + (-3 - 1) \hat{j} = -2 \hat{i} - 4 \hat{j} \]

### Step 3: Find the initial displacement vector from B to A
The initial position of A is \( (4, 5) \) and the position of B is \( (0, -3) \). The displacement vector \( \vec{d_{AB}} \) from B to A is:
\[ \vec{d_{AB}} = (4 - 0) \hat{i} + (5 - (-3)) \hat{j} = 4 \hat{i} + (5 + 3) \hat{j} = 4 \hat{i} + 8 \hat{j} \]

### Step 4: Calculate the time taken to collide
The time taken \( t \) for the particles to collide can be calculated using the formula:
\[ t = \frac{\text{displacement}}{\text{relative velocity}} = \frac{\vec{d_{AB}}}{\vec{v_{AB}}} \]
Calculating the magnitude of \( \vec{v_{AB}} \):
\[ |\vec{v_{AB}}| = \sqrt{(-2)^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \]
Now, the time \( t \) is:
\[ t = \frac{|\vec{d_{AB}}|}{|\vec{v_{AB}}|} = \frac{\sqrt{(4)^2 + (8)^2}}{2\sqrt{5}} = \frac{\sqrt{16 + 64}}{2\sqrt{5}} = \frac{\sqrt{80}}{2\sqrt{5}} = \frac{4\sqrt{5}}{2\sqrt{5}} = 2 \, \text{s} \]

### Step 5: Calculate the x-coordinate of the collision point
The x-coordinate of particle A at time \( t \) is given by:
\[ x_A = x_{A0} + v_{Ax} \cdot t \]
Where \( x_{A0} = 4 \, \text{m} \) and \( v_{Ax} = 2 \, \text{m/s} \):
\[ x_A = 4 + 2 \cdot 2 = 4 + 4 = 8 \, \text{m} \]

### Final Answer
The x-coordinate of the point where the particles collide is \( 8 \, \text{m} \).

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