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 $v_A=2jm{s}^{-1}$ and $v_B=2lm{s}^{-1}$ respectively. The time after which A arid B are at their closest distance is

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

Answer 1

The correct option is A $2.5s$

$y=\overline{0-V_{B}t{}^2+V_{A}t10{}^2}$
or $y^2=(2t{)}^2+(2t-10{)}^2$
or $y^2=1=4t^2+4t^2+100-40t$
$\Rightarrow l=8t^2+10040t$
Now, $\frac{dl}{dt}(16t-40)=0$
$t=\frac{40}{16}2.5s$
As $\frac{d{}^{2}l}{dt{}^2}=-16=(-Ve)$
Hence, $l$ will be minimum.