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) $a /(v+v_1)$
(c) $a /(v-v_1)$
(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,
$AC=v t$ and $BC=v_1 t$
From pythagorus theorem,
$AC^{2}=AB^{2}+BC^{2}$
$v^{2} t^{2}=a^{2}+v_1^{2} t^{2}$
$t^{2}(v^{2}-v_1^{2})=a^{2}$
$t^{2}=\frac{a^{2}}{(v^{2}-v_1^{2})}$
$t=\sqrt{\frac{a}{(v^{2}-v_1^{2})}}$
