A particle moves over three quarters of a circle of radius R what is the magnitude of its displacement
Answers
Answer: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
Explanation:Hope this helps u plz mark me as the brainliest
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