Two boys are standing at the ends A and B of a ground, where `AB=a`. The boy at B starts running in a direction perpendicular to AB with velocity `v_(1)`. The boy at A starts running simultaneously with velocity v and catches the other boy in a time t, where t is :

Options

A. `a/(sqrt(v+v_(1)^(2)))`
B. `sqrt(a/(v^(2)+v_(1)^(2)))`
C. `a/((v-v_(1))`
D. `a/((v+v_(1))`

Explanation

To solve the problem, we need to analyze the motion of both boys and determine the time \( t \) it takes for boy A to catch up with boy B.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- Let the distance between points A and B be \( AB = a \).
- Boy B runs perpendicular to line AB with a velocity \( v_1 \).
- Boy A runs towards boy B with a velocity \( v \).

2. **Position of Boys**:
- At time \( t \), the position of boy B, who started at point B, will be at a distance \( d_B = v_1 \cdot t \) from point B, moving in a direction perpendicular to AB.
- The position of boy A, who started at point A, will be at a distance \( d_A = v \cdot t \) towards point B.

3. **Using Pythagorean Theorem**:
- When boy A catches boy B, the distance between them forms a right triangle where:
- One leg is the distance boy A has covered towards B, which is \( d_A = v \cdot t \).
- The other leg is the distance boy B has covered perpendicular to AB, which is \( d_B = v_1 \cdot t \).
- The hypotenuse of this right triangle is the distance \( AB = a \).

4. **Setting Up the Equation**:
- According to the Pythagorean theorem:
\[ (v \cdot t)^2 + (v_1 \cdot t)^2 = a^2 \]
- Simplifying this gives:
\[ v^2 t^2 + v_1^2 t^2 = a^2 \]
- Factoring out \( t^2 \):
\[ t^2 (v^2 + v_1^2) = a^2 \]

5. **Solving for Time \( t \)**:
- Rearranging the equation to solve for \( t^2 \):
\[ t^2 = \frac{a^2}{v^2 + v_1^2} \]
- Taking the square root of both sides to find \( t \):
\[ t = \frac{a}{\sqrt{v^2 + v_1^2}} \]

### Final Answer:
The time \( t \) it takes for boy A to catch boy B is given by:
\[ t = \frac{a}{\sqrt{v^2 + v_1^2}} \]