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Two persons A and B are located in X-Y plane at the points at the points (0,0) and (0,10) respectively. (The distance are measured in MKS units). At a time $t=0$, they start moving simultaneously with velocities $\bar {v}_A=2 \hat j m/s$ and $\bar v_B=2 \hat i m/s$ respectively. The time after which A and B are at their closest distance is :

\[\begin {array} {1 1} (1)\;2.5\;sec & \quad (2)\;4\;sec \\ (3)\;1\;sec & \quad (4)\;\frac{10}{\sqrt 2}\;sec \end {array}\]
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1 Answer

(1) 2.5 sec
answered Nov 7, 2013 by pady_1
 

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