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=2 j ms ⁻¹ and v _B=2 i ms ⁻¹ respectively. The time after which A and B are at their closest distance is

Step-by-step solution

Let after the time (t) the position of A is (0, v_A t ) and position of B= (v_B t, 10 ) . Distance between them aligned y & = (0-v_B t )^2+ (v_A t-10 )^2 or y^2 & =(2 t)^2+(2 t-10)^2 aligned or y^2=l=4 t^2+4 t^2+100-40 t l=8 t^2+100-40 t Now, d l d t =(16 t-40)=0t= 40 16 =2.5 ~s As d^2 l d t^2 =16=(+ ve ) Hence, l will be minimum.