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Displacement and distance

  1. Jun 1, 2014 #1
    I am quite confused between displacement and distance .

    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.
     
  2. jcsd
  3. Jun 1, 2014 #2

    Matterwave

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    There's really no need to do integrals for this problem. Really the problem is the understanding of the difference between displacement and distance. Distance only cares about the total amount of moving that you do, without regard for direction. Displacement cares about the direction in which you are moving.

    The difference is exactly analogous to speed and velocity. Speed cares only about the absolute quickness with which you move, while velocity cares about the direction.

    Therefore, if I walked 5 steps to the right, and then walked 5 steps to the left, ending up where I started, my distance traveled is 10 steps (just add all the motion I was involved in), while my displacement is 0. This is how these concepts are defined.

    As for your example, the distance traveled is along the circle. You add up the entire length of the arc of the circle that you traveled since distance doesn't care about direction. In this case you get ##3\pi## meters which should be easy enough to obtain without integrating anything. The displacement, which you noted in your post, is simply the difference between the final position and the initial position because now I DO care about directions! In this case, it's also quite simple to see that the displacement is -6i meters.

    SO even though I've walked a DISTANCE of ##3\pi\approx 9.28m>6m##, I've only made a DISPLACEMENT of 6m to the left because some of my motion (namely the vertical motion that I've made) has canceled out from the two sides of the arc.
     
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