Two particles, A and B, are in motion in the xy-plane. Their co-ordinates at each instant of time `t (t ge 0) ` are given by `x_(A) =t, y _(A)=2t , x _(B)=1-t and y _(B) =t.` The minimum distance between particles A and B is :

Options

A. `1/5`
B. `(1)/(sqrt5)`
C. 1
D. `sqrt((2)/(3))`

Answer

To find the minimum distance between the two particles A and B with given coordinates, we can follow these steps:

### Step 1: Write down the coordinates of the particles
The coordinates of the particles A and B at time \( t \) are given as:
- Particle A: \( (x_A, y_A) = (t, 2t) \)
- Particle B: \( (x_B, y_B) = (1 - t, t) \)

### Step 2: Use the distance formula
The distance \( d \) between the two particles can be calculated using the distance formula:
\[ d = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} \]

### Step 3: Substitute the coordinates into the distance formula
Substituting the coordinates into the distance formula, we have:
\[ d = \sqrt{((1 - t) - t)^2 + (t - 2t)^2} \]
This simplifies to:
\[ d = \sqrt{(1 - 2t)^2 + (-t)^2} \]

### Step 4: Simplify the expression
Now, we simplify the expression inside the square root:
\[ d = \sqrt{(1 - 2t)^2 + t^2} \]
Expanding this gives:
\[ d = \sqrt{(1 - 4t + 4t^2) + t^2} = \sqrt{5t^2 - 4t + 1} \]

### Step 5: Minimize the quadratic expression
To find the minimum distance, we need to minimize the expression \( 5t^2 - 4t + 1 \). This is a quadratic equation in the form \( at^2 + bt + c \), where \( a = 5 \), \( b = -4 \), and \( c = 1 \).

### Step 6: Find the vertex of the quadratic
The minimum value of a quadratic \( at^2 + bt + c \) occurs at \( t = -\frac{b}{2a} \):
\[ t = -\frac{-4}{2 \cdot 5} = \frac{4}{10} = \frac{2}{5} \]

### Step 7: Substitute \( t \) back into the quadratic
Now, substitute \( t = \frac{2}{5} \) back into the quadratic to find the minimum value:
\[ q = 5\left(\frac{2}{5}\right)^2 - 4\left(\frac{2}{5}\right) + 1 \]
Calculating this gives:
\[ q = 5 \cdot \frac{4}{25} - \frac{8}{5} + 1 = \frac{20}{25} - \frac{40}{25} + \frac{25}{25} = \frac{20 - 40 + 25}{25} = \frac{5}{25} = \frac{1}{5} \]

### Step 8: Find the minimum distance
The minimum distance \( d \) is then:
\[ d_{\text{min}} = \sqrt{q_{\text{min}}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} \]

### Final Answer
The minimum distance between particles A and B is \( \frac{1}{\sqrt{5}} \).

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