Asked by | 17th Apr, 2012, 09:55: PM
Expert Answer:
angle ABC=60degrees (equilateral triangle)
component of velocity of B along AB =vcos60=v/2
A moves towards B with velocity 'v'
Therefore rate of decrease of the distance 'a' between A and B is=v+v/2=3v/2
time required for distance between A & B from a to zero is =a/(3v/2)=2a/3v
angle ABC=60degrees (equilateral triangle)
component of velocity of B along AB =vcos60=v/2
A moves towards B with velocity 'v'
Therefore rate of decrease of the distance 'a' between A and B is=v+v/2=3v/2
time required for distance between A & B from a to zero is =a/(3v/2)=2a/3v
Answered by | 18th Apr, 2012, 01:30: PM
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