Question

Two persons $A$ and $B$ are located in $X-Y$ plane at the points $(0,0)$ and $(0,10)$, respectively. (The distance are measured in $MKS$ units). At a time $t=0$, they start moving simultaneously with velocities ${\overrightarrow{v}}_A=2\hat{j}m{s}^{-1}$ and ${\overrightarrow{v}}_B=2\hat{i}m{s}^{-1}$, respectively. The time after which $A$ and $B$ are their closest distance is

• A. $2.5s$
• B. $4s$
• C. $1s$
• D. $\frac{10}{\sqrt{2}}s$

Answer 1

The correct option is A $2.5s$

Given,

Initial separation in both $10m$.

$v_A=\frac{dx}{dt}=2m{s}^{-1}$ (person A movement increase value of x).

$v_B=\frac{-dy}{dt}=2m{s}^{-1}$ (person B movement decrease value of Y).

Distance between both persons is $s$

$s^2=x^2+y^2$

Differentiate with respect to time

$2sds=2xdx+2ydy$

For minimum, approach first differentiation is zero.

$2xdx+2ydy=0$

$\Rightarrow \frac{x}{y}=\frac{-dy}{dx}=\frac{-dy/dt}{dx/dt}=\frac{2}{2}=1$

$\Rightarrow x=y$

$\Rightarrow 2t=10-2t$

$\Rightarrow t=2.5\sec$

$x=2t=2\times 2.5=5m$

$x=y=5m$

$s=\sqrt{x^2+y^2}=\sqrt{5^2+5^5}=5\sqrt{5}m$

Minimum approach is $5\sqrt{5}m$, In time $2.5\sec$