SATHEE: Motion in a Plane - Result Question 46

Motion in a Plane - Result Question 46

49. Two boys are standing at the ends $A$ and $B$ of a ground where $A B = a$ . The boy at $B$ starts running in a direction perpendicular to $A B$ 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

[2005]

(a) $a / \sqrt{v^{2} + v_{1}^{2}}$

(b) $a / (v + v_{1})$

(d) $\sqrt{a^{2} / (v^{2} - v_{1}^{2})}$

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Answer:

Correct Answer: 49. (d)

Solution:

(d) According to the given condition,

Let both $y$ boy meet at $c$ : in time $t$ then,

$A C = v t$ and $B C = v_{1} t$

From pythagorus theorem,

$A C^{2} = A B^{2} + B C^{2}$

$v^{2} t^{2} = a^{2} + v_{1}^{2} t^{2}$

$t^{2} (v^{2} - v_{1}^{2}) = a^{2}$

$t^{2} = \frac{a^{2}}{(v^{2} - v_{1}^{2})}$

$t = \sqrt{\frac{a}{(v^{2} - v_{1}^{2})}}$

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Motion in a Plane - Result Question 46

49. 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 v1. The boy at A starts running simultaneously with velocity v and catches the other boy in a time t, where t is

Answer:

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49. Two boys are standing at the ends $A$ and $B$ of a ground where $A B = a$ . The boy at $B$ starts running in a direction perpendicular to $A B$ 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