a particle moves over 3 quarters of circle of radius 7cm what is the distance covered and magnitude of displacement
Answers
Answer:
Net displacement is sqrt(7^2+7^2), or sqrt(98).
Distance traveled is 3*pi*14/4
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Answer:
Explanation:Here I am assuming that you mean by particle the classical point object. Displacement is a vector quantity that only depends on the initial and final positions of the particle (It is independent of the path taken by the particle). If we have that definition in mind, we define the initial position of the particle in a cartesian coordinate system with the coordinates A(7,0), then we allow the particle to move anti-clockwise along a circle of radius 7 and centered on the origin. When the particle covers three quarters of the circumference of the circle it rests on the point B(0,-7).
The displacement vector x⟶ has coordinates AB⟶ (0–7,-7–0) which is just AB⟶ (-7,-7). It is important to keep in mind that the displacement vector is dependent upon the chosen coordinate system, which means if I choose another coordinate system, the coordinates of x⟶ will change accordingly. Also, changing the direction of rotation from anti-clockwise to clockwise will change the vector coordinates. What’s useful, however, is that the magnitude of the displacement remains unchanged . This magnitude is given by the pythagorean theorem and can be written as follows:
AB2=x2AB+y2AB
Where x2AB is the square of the x-coordinate (abscissa) of AB⟶ and y2AB is the square of the y-coordinate (ordinate) of AB⟶ .
If we perform the calculations, the magnitude of the displacement will be:
72+72−−−−−−√
Which is just 72–√cm .
Conclusion: ∥x⃗ ∥=72–√cm .