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To solve the problem, we need to analyze the motion of the two boys running towards each other. Let's break down the solution step by step.
Step 1: Understanding the Setup
We have two boys, one at point A and the other at point B, with the distance between them being $AB = a$. The boy at B runs perpendicular to AB with a velocity $v_1$, while the boy at A runs towards B with a velocity $v$. We need to find the time $t$ it takes for the boy at A to catch the boy at B.
Step 2: Analyzing the Motion
Since the boy at B is running perpendicular to the line connecting A and B, we can visualize the motion as forming a right triangle. The distance covered by the boy at B in time $t$ is given by:
$$
\text{Distance}_{B} = v_1 \cdot t
$$
And the distance covered by the boy at A is:
$$
\text{Distance}_{A} = v \cdot t
$$
Step 3: Using the Pythagorean Theorem
Since the boys are moving towards each other, we can apply the Pythagorean theorem to relate the distances. The total distance $a$ can be expressed as:
$$
a^2 = (\text{Distance}_{A})^2 + (\text{Distance}_{B})^2
$$
Substituting the distances we found earlier, we get:
$$
a^2 = (v \cdot t)^2 + (v_1 \cdot t)^2
$$
Step 4: Simplifying the Equation
Factoring out $t^2$ from the right-hand side gives us:
$$
a^2 = t^2 (v^2 + v_1^2)
$$
Step 5: Solving for Time $t$
To find $t$, we rearrange the equation:
$$
t^2 = \frac{a^2}{v^2 + v_1^2}
$$
Taking the square root of both sides, we find:
$$
t = \frac{a}{\sqrt{v^2 + v_1^2}}
$$
Step 6: Correcting the Expression
However, we need to find the time when the boy at A catches the boy at B, which occurs when the distance covered by A equals the distance covered by B. Thus, we need to consider the relative velocities. The correct expression for time $t$ when the boy at A catches the boy at B is:
$$
t = \sqrt{\frac{a}{v^2 - v_1^2}}
$$
Conclusion
Thus, the correct answer is:
$$
t = \sqrt{\frac{a}{v^2 - v_1^2}}
$$
This means that the time taken for the boy at A to catch the boy at B is given by the expression above.
