3 persons are initially at the 3 corners of an equilateral triangle

3 persons are initially at the 3 corners of an equilateral triangle whose side is equal to d. Each person now moves with a uniform speed v in such a way that the first moves directly towards the second and second directly towards the 3rd and 3rd directly towards the first. 3 persons will meet after a time equal to

Options

(a) d/v s
(b) 2d/3v s
(c) 2d/v√3 s
(d) d/v√3 s

Correct Answer:

2d/3v s

Explanation:

No explanation available. Be the first to write the explanation for this question by commenting below.

admin:

View Comments (1)

  • D=UT+1/2AT2
    U=0 (GIVEN)
    D=1/2AT2
    2D/A=T2

    V2=U2+2AS (3PERSONS)
    (U=0)
    (3V)2==2AS
    9V2/2D=A
    2D/A=T2
    2D*2D/9V2==T2
    T==2D/3V

Related Questions

  1. The difference in the lengths of a mean solar day and a sidereal day is about
  2. A dust packet is dropped from 5th storey of a multi-storeyed building
  3. A beam of light of λ=600 nm from a distant source falls on a single slit 1 mm wide
  4. When a ceiling fan is switched on, it makes 10 revolutions in the first 3s
  5. In which case positive work has to be done in seperating them?