Two particles are moving along two long straight lines, in the same plane with same speed equal to `20 cm//s.` The angle between the two linse is `60^@` and their intersection point isO. At a certain moment, the two particles are located at distances 3m and 4m from O and are moving twowards O. Subsequently, the shortest distance between them will be

Options

A. `50 cm`
B. `40sqrt2 cm`
C. `50sqrt2 cm`
D. `50sqrt3 cm`

Explanation

To solve the problem step-by-step, we will analyze the motion of the two particles and find the shortest distance between them as they move towards the intersection point O.

### Step 1: Understand the Problem
We have two particles moving towards a common point O along two lines that intersect at an angle of 60 degrees. Particle P is initially 3 meters from O, and particle Q is 4 meters from O. Both particles are moving at the same speed of 20 cm/s.

### Step 2: Convert Distances to Centimeters
Since the speed is given in cm/s, we should convert the distances of the particles from meters to centimeters:
- Distance of P from O = 3 m = 300 cm
- Distance of Q from O = 4 m = 400 cm

### Step 3: Set Up the Coordinate System
We can set up a coordinate system where:
- The x-axis represents the line along which particle P is moving.
- The y-axis represents the line along which particle Q is moving.

### Step 4: Represent the Velocities
- The velocity of particle P (moving towards O) can be represented as:
\[ \vec{V_P} = -20 \hat{i} \text{ cm/s} \]
- The velocity of particle Q can be resolved into components based on the angle of 60 degrees:
\[ \vec{V_Q} = -20 \cos(60^\circ) \hat{i} - 20 \sin(60^\circ) \hat{j} \]
\[ \vec{V_Q} = -10 \hat{i} - 10\sqrt{3} \hat{j} \text{ cm/s} \]

### Step 5: Calculate the Relative Velocity
To find the shortest distance, we will consider the relative motion of Q with respect to P. The relative velocity \(\vec{V_{QP}}\) is given by:
\[ \vec{V_{QP}} = \vec{V_Q} - \vec{V_P} \]
Substituting the values:
\[ \vec{V_{QP}} = (-10 \hat{i} - 10\sqrt{3} \hat{j}) - (-20 \hat{i}) = 10 \hat{i} - 10\sqrt{3} \hat{j} \]

### Step 6: Determine the Direction of the Relative Velocity
To find the angle of the relative velocity, we can use:
\[ \tan(\theta) = \frac{\text{j-component}}{\text{i-component}} = \frac{-10\sqrt{3}}{10} = -\sqrt{3} \]
This indicates that the angle is 60 degrees below the x-axis.

### Step 7: Calculate the Shortest Distance
The shortest distance will occur along the line perpendicular to the direction of the relative velocity. We can find the length of this perpendicular line using trigonometric relationships.

Let \(A\) be the point where the shortest distance occurs. The distance \(AM\) can be calculated as follows:
- The distance \(AM\) can be determined using the sine of the angle:
\[ \sin(60^\circ) = \frac{QM}{PM} \]
Where \(PM\) is the distance from P to O (300 cm) and \(QM\) is the distance from Q to O (400 cm).

Using the Pythagorean theorem:
\[ PM^2 + QM^2 = PQ^2 \]
\[ 300^2 + 400^2 = PQ^2 \]
\[ PQ = 500 \text{ cm} \]

Now, the shortest distance \(d\) can be calculated as:
\[ d = AM \cdot \sin(60^\circ) = 500 \cdot \frac{\sqrt{3}}{2} = 250\sqrt{3} \text{ cm} \]

### Final Answer
The shortest distance between the two particles as they move towards O is:
\[ \text{Shortest Distance} = 250\sqrt{3} \text{ cm} \]