Question

A particle moves three quarters of a circle of radius r. What is the magnitude of its displacement

Answer 1

Answer:

Consider the figure below

here the particle has moved from point A to point B in an anticlockwise direction (shown by arrows). The net displacement in this case will be equal to AB (distance between initial and final position).

now, in triangle AOB, according to the pythagoras theorem

AB2 = OA2 +OB2

or

AB2 = r2 + r2

thus, the displacement will be

AB = r√2

Answer 2

Given particle covers three quarters of a circle with radius R.

Distance travelled = 3/4(circumference of circle)

=3/4(2*pi* r)

= 3/2 ( pi* r).

DISPLACEMENT=SHORTEST PATH

= IF WE JOIN POINT A and B we get right angle triangle with

Hypotenuse(displacement) and other two sides of triangle with sides r.

Displacement =Sqr(r^2 +r^2)

=sqr(2r^2)

=r*sqr(2)

=1.41*r meters

Therefore Displacement covered by particle is 1.414*r m