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Solution of task using graph

December 29th, 2011 . Posted in Classical physicsNo Comments »
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   The process of solving the problem is a type of mental activity, so to successfully meet the challenges need to be clear about the logical relationships between the physical quantities. To build a system of logical connections let’s consider a mathematical object that has the name of a graph.

     To get an idea of the graph let’s consider an example of the problem of the bridges of the city of Konigsberg. Konigsberg (now Kaliningrad) consisted of 4 parts joined using 7th bridges. The Swiss mathematician Leonhard Euler in 1736 proved that it is impossible to go through all the bridges without going on one of them twice. To prove this fact for the first time in history he built a graph whose vertices are designated part of the city and connecting their arch-bridges.

     Graphs are widely used in mathematics, computer science, cognitive linguistics, and expert systems. Let’s apply the graph for representing knowledge in physics and solving physics problems with it. The vertices of the graph, in this case, will be the physical quantity, and arcs – integral and differential relations, by which to move from one to another physical quantity. Let’s consider the process of constructing an elementary graph that displays the differential relationship between the displacement and velocity of the material point:

 \vec v(t) = \frac {d \vec s(t)} {dt}

Let’s locate in the vertices the displacement  \vec s и скорость \vec v, and denote the arc as derivative, as shown below:

My_site_Irodov

Let’s apply this graph to solve the problem of the number 1.25b from the collection I.E. Irodov “Problems in General Physics” .- M .: Nauka.-1988

Condition of the Task: The point moves in the XOY plane according to the law: x = {\alpha} t y = {\alpha} t (1-{\beta}t) . Find the dependence of the speed of point v of the time.

Solution. The equation of motion is given in the parametric form, so the transition from the vertice \vec s the vertice\vec v  will take partial derivatives, which will give the vector components \vec v: v_x  и v_y.

v_x = \frac{dx} {dt}=\alpha, v_y = \frac{dy }{dt} = {\alpha}(1-{\beta}t)+{\alpha} t(-{\beta}) = {\alpha} (1-2{\beta}t)

Knowing the components of the vector  \vec v along the axes, we find the dependence of the velocity in terms of time:

v = \sqrt {v_x^2 + v_y^2} = \sqrt {{\alpha}^2 + {\alpha}^2 (1-2{\beta}t)^2} = {\alpha} \sqrt {1 + (1-2{\beta}t)^2 }

   We have constructed an elementary graph that does not show all the relationships between the physical quantities in mechanics.

       To learn how to build a complete graph for the representation of knowledge in kinematics and apply it to solving problems, you can find out by signing up for our free newsletter.

(More information can be found at :

https://ipood-kiev.academia.edu/ValentynaShvets)

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