Three particles A, B and C are situated at the vertices of an equilateral triangle ABC of side d at time t = 0. Each of the particles moves with constant speed v. A always has its velocity along AB, B along BC and C along CA. At what time will the particles meet each other ?
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- Don't want spam or copied answers
- Don't be greedy for points
- Best answer will be awarded with brainliest.
Answers & Comments
Answer:
Velocity of A is v along AB. The velocity of B is along BC.Its component
along BA is vcos60=
2
v
. Thus, the separation AB decreases at the rate
v−(−
2
v
)=
2
3v
Since, this rate is constant, the time taken in reducing the separation AB from d to zero is
t=
(3v/2)
d
=
3v
2d
Verified answer
Answer:
Velocity of A is v along AB. The velocity of B is along BC.Its component
along BA is vcos60= v/2 Thus, the separation AB decreases at the rate
v−(− v/2) =3v/2
Since, this rate is constant, the time taken in reducing the separation AB from d to zero is
t= d/(3v/2) = 2d/3v
Explanation:
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