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}=2 j \,ms ^{-1} $ and $\overrightarrow{ v }_{B}=2 \hat{ i } \,ms ^{-1}$ 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 $y^{2}=(2\, t)^{2}+(2\, t-10)^{2}$
or $y^{2}=l=4\, t^{2}+4 \,t^{2}+100-40 \,t$
$\Rightarrow l =8\, t^{2}+100-40 \,t $
Now,$ \frac{d l}{d t} =(16\, t-40)=0 $
$ t =\frac{40}{16}=2.5 \,s $
As $\frac{d^{2} l}{d t^{2}}=16=(+v e)$
Hence, $l$ will be minimum.