Question

Two particles P and Q move with constant velocities v1 = 2 ms" and v2 = 4 ms along two mutually perpendicular straight lines towards t1
point 0 At moment t = 0, the particles were located at distances l1 = 12 m and I 2 = 19 m from O, respectively. Find the time in second) when they are nearest​

Answer 1

The situation is shown of figure. After time t let separation between them is x, then

x2=(l1−v1t)2+(l2−v2t)2.....(i)

x to be minimum, dtdx=0

Differentiating equation (i) w.r.t time

we have dtdx2=dtd[(l1−v1t)2+(l2−v2r)2]

2xdtdx=2(l1−v1t)×(−v1)+2(l1−v2t)×(−v2)

or 0=(l1−v1t)(v1+(l2−v2t)v2=(l1v1−l2v2)−t(v12+v22)

which gives t=(v12+v22l1v1+l2v2)

Now xmin=(l1−v1t)2+(l2−v2t)2.....(ii)

Substituting value of t in equation (ii), we get

xmin=

[l1−v1(v12+v22l1v1+l2v2)]2+[l1−v2(v12+v22l1v1−l2v2)]2

After solving, we get xmin=(v12+v22l1v2−l2v1).

Explanation:

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