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. `a^(2)y+bx^(2)=0`
B. `a^(2)y=bx^(2)`
C. `y=b/a^(2) x`
D. `ay^(2)=b^(2)x`

Explanation

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

### Step 1: Determine the coordinates of particle A at \( t = 1 \)

The position of particle A is given by:
\[ x_A(t) = 3t \]
\[ y_A(t) = t \]

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:
\[ x_B(t) = 6 \]
\[ y_B(t) = 2 + 3t^2 \]

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: Use the distance formula to find 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:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Substituting the coordinates of A and B:
\[ d = \sqrt{(6 - 3)^2 + (5 - 1)^2} \]

Calculating the differences:
\[ d = \sqrt{(3)^2 + (4)^2} \]

Calculating the squares:
\[ d = \sqrt{9 + 16} \]

Adding the results:
\[ d = \sqrt{25} \]

Finally, taking the square root:
\[ d = 5 \]

### Conclusion

The distance between the two particles A and B at \( t = 1 \) second is \( 5 \) units.

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