Physics, asked by kittykat7, 1 year ago

a particle moves over three quarters of a circle of radius r .what is the magnitude of its displacement ?

Answers

Answered by prmkulk1978
815
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
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Answered by vkpk
508
Given :

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

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