Question

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

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

Answer 1

Answer: (A)

Let after the time $(t)$ the position of $A$ is $\left(0,v_{A}t\right)$ and position of $B=\left(v_{B}t,10\right)$. Distance between them
$y=\sqrt{{\left(0-v_{B}t\right)}^2+{\left(v_{A}t-10\right)}^2}$
or $y^2=(2t{)}^2+(2t-10{)}^2$
or $y^2=l=4t^2+4t^2+100-40t$
$\Rightarrow l=8t^2+100-40t$
Now,$\frac{dl}{dt}=(16t-40)=0$
$t=\frac{40}{16}=2.5s$
As $\frac{d^{2}l}{d{t}^2}=16=(+ve)$
Hence, $l$ will be minimum.