Question

a particle moves over three quarters of a circle of radius of 7m.what is distance covered and magnitude of displacement

Answer 1

hey friend!!

_______-

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 = 1.41×7 = 9.87 meter___ans

Therefore Displacement covered by particle is " 9.87 meter"

hope it will help you

Answer 2

As we know that displacement is the shortest path between source and destination.
Hence if we join the point A and B we get a Right angled triangle.
where
(displacement)=Hypotenuse
 and other two sides of triangle with sides r.
Displacement =√(r² +r²)
=√(2r²)
=r*√(2)
=1.41*r meters
∴ Displacement covered by particle is 1.414*r m