# Relative Velcity: Two particals moing in different direction

Two particles leave the same point at time t=0, each one over a straight line with constant acceleration a. If theta is the angle of the two lines, how varies the distance d between the particles with time?

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Relative Motion

Two particles leave the same point at time t=0, each one over a straight line with constant acceleration a. If Ó¨ is the angle of the two lines, how varies the distance d between the particles with time?

Solution:

The particles leaves the same point O with uniform acceleration of magnitude a, along lines OA and OB

Initial velocity of both the particles is u = 0

Then the magnitude of displacement s of each particle is given by using second equation of motion as

[s = u t + Â½ at2]

s = 0 + Â½ at2

or s = Â½ at2 --------------------- (1)

Now as the angle between the two lines of motion is Î¸, the distance between the two particles at time t is given by the length of the line joining the two particles at time t which is AB.

Now drawing perpendicular OM from point O on line AB, as the two sides OA and OB are equal, it will bisect AB as well as the angle Î¸.

Using the property of a right angled triangles we have

Gives

Hence the distance between the two particles will be

Substituting from equation (1) we get

Or

Alternate solution:

Let the acceleration of one particle is and that of the other is

Then the acceleration of the second particle relative to the first particle is given by

As the magnitude of both accelerations is same 'a' and the angle between the directions is Î¸, thus the magnitude of vector is given by using the result from the law of parallelogram of vectors as

Or

Or

Or

Hence the distance of second particle from the first after time t is given by

d = 0t + Â½ aRt2 = Â½ {2 a sin (Î¸/2)} t2

or d = at2 sin (Î¸/2).

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