Question

Three particles are initially at the vertices of an equilateral triangle. The particle A and B start
to move with constant velocities va and vb in any direction. Which relation satisfies
the speed of the 3rd particle C such that all particles lie on equilateral triangle ?​

Answer 1

Answer:

Each of the particles moves with constant speed v. ... Three particles A, B and C are situated at the vertices of an equilateral triangle ... The velocity of B is along BC. ... No Signup required.

FOLLOW ME dear friend

Answer 2

Answer:

Explanation:

Concept:

Definition of distance:

Distance is the sum of an object's movements, regardless of direction. Distance can be defined as the amount of space an object has covered, regardless of its starting or ending position.

Definition of speed:

Speed is define as the speed at which an object's location changes in any direction. The distance travelled in relation to the time it took to travel that distance is how speed is defined. Since speed simply has a direction and no magnitude, it is a scalar quantity.

Given:

vertices of an equilateral triangle

constant velocities

To find:

speed

Solution:

1st approach:

The velocity of (A) is along in 2 (a) 45. (AB) (B) is moving with an along (BC) The velocity of (B) along BA is given by  $vcos60=v/2.$ As a result, the distance between the parts at (A) and (B) decreases at a rate of speed= $v + v/2$

         = $3v/2.$

Given that this speed is constant, shrinking the spacing will take time.

         = $2a/3v$

2nd approach:

$AO = v sqrt 3 /2'@ = v cos 30$

distance AO=$2/3AD$

                   =$2/3\sqrt(a^ 2 - a^ 2/4)$

                   = $a/\sqrt 3$

Consequently, the time needed for a particle at (A) to travel from (A) to (O).

$AO/v cos30 =a/\sqrt 3/v\sqrt 3/2$

                  $= 2a/3v$

#SPJ3