Two persons A and B are located in the X-Y pl | vedclass

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(A) Let at time $t$,the position of $A$ be $\overrightarrow{r}_A = (0,0) + (0,2)t = (0, 2t)$.Let at time $t$,the position of $B$ be $\overrightarrow{r

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$2.5 	ext{ s}$

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(A) Let at time $t$,the position of $A$ be $\overrightarrow{r}_A = (0,0) + (0,2)t = (0, 2t)$.Let at time $t$,the position of $B$ be $\overrightarrow{r}_B = (0,10) + (2,0)t = (2t, 10)$.The displacement vector between them is $\overrightarrow{r}_{BA} = \overrightarrow{r}_B - \overrightarrow{r}_A = (2t - 0, 10 - 2t) = (2t, 10 - 2t)$.The square of the distance $D^2$ between them is given by $f(t) = (2t)^2 + (10 - 2t)^2$.$f(t) = 4t^2 + 100 + 4t^2 - 40t = 8t^2 - 40t + 100$.To find the time for the minimum distance,we differentiate $f(t)$ with respect to $t$ and set it to zero:$\frac{df}{dt} = 16t - 40 = 0$.$16t = 40 \implies t = \frac{40}{16} = 2.5 	ext{ s}$.Since the second derivative $\frac{d^2f}{dt^2} = 16 > 0$,the distance is indeed minimum at $t = 2.5 	ext{ s}$.

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