Two particle A and B are moving in XY-plane. Their positions vary with time t according to relation
`x_(A)(t)=3t, x_(B)(t)=6`
`y_(A)(t)=t, y_(B)(t)=2+3t^(2)`
The distance between two particle at `t=1` is :

Options

A. 5
B. 3
C. 4
D. `sqrt(12)`

Explanation

To find the distance between two particles A and B at time \( t = 1 \), we will follow these steps:

### Step 1: Determine the coordinates of particle A at \( t = 1 \)
The position of particle A is given by:
\[ x_A(t) = 3t \quad \text{and} \quad y_A(t) = t \]
Substituting \( t = 1 \):
\[ x_A(1) = 3 \times 1 = 3 \]
\[ y_A(1) = 1 \]
Thus, the coordinates of particle A at \( t = 1 \) are \( (3, 1) \).

### Step 2: Determine the coordinates of particle B at \( t = 1 \)
The position of particle B is given by:
\[ x_B(t) = 6 \quad \text{and} \quad y_B(t) = 2 + 3t^2 \]
Substituting \( t = 1 \):
\[ x_B(1) = 6 \]
\[ y_B(1) = 2 + 3 \times 1^2 = 2 + 3 = 5 \]
Thus, the coordinates of particle B at \( t = 1 \) are \( (6, 5) \).

### Step 3: Calculate the distance between particles A and B
The distance \( d \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by the formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Here, we have:
- \( (x_1, y_1) = (3, 1) \) for particle A
- \( (x_2, y_2) = (6, 5) \) for particle B

Substituting the coordinates into the distance formula:
\[ d = \sqrt{(6 - 3)^2 + (5 - 1)^2} \]
Calculating each term:
\[ d = \sqrt{(3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

### Final Answer
The distance between the two particles at \( t = 1 \) is \( 5 \) units.

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