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

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$2.5$ s

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(A) Let at time $t$, the position of $A$ be $\overrightarrow{r}_A = (0 + 0t)\hat{i} + (0 + 2t)\hat{j} = 2t\hat{j}$.Let at time $t$, the position of $B$ be $\overrightarrow{r}_B = (0 + 2t)\hat{i} + (10 + 0t)\hat{j} = 2t\hat{i} + 10\hat{j}$.The relative position vector $\overrightarrow{r}_{AB} = \overrightarrow{r}_B - \overrightarrow{r}_A = 2t\hat{i} + (10 - 2t)\hat{j}$.The square of the distance $D^2 = |\overrightarrow{r}_{AB}|^2 = (2t)^2 + (10 - 2t)^2$.$D^2 = 4t^2 + 100 + 4t^2 - 40t = 8t^2 - 40t + 100$.For minimum distance, $\frac{d(D^2)}{dt} = 0$.$16t - 40 = 0$.$t = \frac{40}{16} = 2.5 	ext{ s}$.Since $\frac{d^2(D^2)}{dt^2} = 16 > 0$, the distance is minimum at $t = 2.5 	ext{ s}$.

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