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Question

Three particles start simultaneously from a point on a horizontal smooth plane. First particle moves with speed v1 towards east, second particle moves towards north with speed v2 and third-one moves towards north east. The velocity of the third particle, so that the three always lie on a line, is

A
v1+v22
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B
v1v2
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C
v1v2v1+v2
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D
2v1v2v1+v2
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Solution

The correct option is D 2v1v2v1+v2
Equation of line RS is y=mx+C


y=(v2v1)x+v2t
v1y=v2x+v1v2t
Equation of line OP is
y=x

v1x=v2x+v1v2t
Point P is the point of intersection we get,
xp=yp=v1v2tv1+v2
OP=x2p+y2p=2v1v2tv1+v2
Hence, velocity of third particle is
v3=OPt=2v1v2v1+v2

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