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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 $\quad \overrightarrow{\mathbf{v}}_A=2 \mathbf{j} \mathrm{ms}^{-1} \quad$ and $\quad \overrightarrow{\mathbf{v}}_B=2 \hat{\mathbf{i}} \mathrm{ms}^{-1}$ respectively. The time after which $A$ and $B$ are at their closest distance is
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The correct answer is:
4 s
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
$\begin{aligned} y & =\sqrt{\left(0-v_B t\right)^2+\left(v_A t-10\right)^2} \\ \text { or } \quad y^2 & =(2 t)^2+(2 t-10)^2\end{aligned}$
or $\quad y^2=l=4 t^2+4 t^2+100-40 t$ $\Rightarrow \quad l=8 t^2+100-40 t$
Now, $\quad \frac{d l}{d t}=(16 t-40)=0$
$t=\frac{40}{16}=2.5 \mathrm{~s}$
As $\frac{d^2 l}{d t^2}=16=(+\mathrm{ve})$
Hence, $l$ will be minimum.
$\begin{aligned} y & =\sqrt{\left(0-v_B t\right)^2+\left(v_A t-10\right)^2} \\ \text { or } \quad y^2 & =(2 t)^2+(2 t-10)^2\end{aligned}$
or $\quad y^2=l=4 t^2+4 t^2+100-40 t$ $\Rightarrow \quad l=8 t^2+100-40 t$
Now, $\quad \frac{d l}{d t}=(16 t-40)=0$
$t=\frac{40}{16}=2.5 \mathrm{~s}$
As $\frac{d^2 l}{d t^2}=16=(+\mathrm{ve})$
Hence, $l$ will be minimum.
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