I am quite confused between displacement and distance .(adsbygoogle = window.adsbygoogle || []).push({});

Suppose a particle travels along a semi circular path or radius 3 m with a constant speed 5 m/s starting from one end of the diameter A and ending at the opposite end B . Origin is at the center of the circular path.

Case 1) Let the particle travel a distance ds in time dt .Then ##ds=3dθ## →##\int ds=\int_0^\pi 3dθ## → ## s = 3\pi ## .Hence distance travelled ## s = 3\pi##

Case 2) Let the particle travel a distance ds in time dt .Then ##ds=vdt## → ##\int ds=\int_0^t 5dt## → ##\int ds = 5\int_0^{\frac{3\pi}{5}}dt## .Hence distance travelled ## s = 3\pi##

Case 3) The position vector of the particle is given by $$ \vec {r} = 3cosθ\hat{i}+3sinθ\hat{j}$$

$$ \frac{\vec {dr}}{dt} = -3sinθ\frac{dθ}{dt}\hat{i}+3cosθ\frac{dθ}{dt}\hat{j} $$

$$ \vec{v} = -3sinθ\frac{dθ}{dt}\hat{i}+3cosθ\frac{dθ}{dt}\hat{j} $$

$$ \int \vec{dr} = \int \vec{v}dt $$

$$ \int \vec{dr} = \int_0^\pi-3sinθdθ\hat{i}+\int_0^\pi3cosθdθ\hat{j} $$

$$ \int \vec{dr} = -6\hat{i}$$ which is vector difference between final and initial position vectors.

My doubts are

1) Why is ##ds## different from ##\vec{dr}## .Shouldn't ## \int \vec{dr}## and ## \int ds## ,both give distance between the initial and final position .

2) What exactly does ## \int \vec{dr}## tell us ?

3) Can we not calculate distance traveled by a body by using kinematics equations ,which invariably give displacement ?

Please help me in understanding these concepts.

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# Displacement and distance

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