In tasks on addition of speeds the motion of bodies happens, as a rule, uniform and rectilinear and is described by the simple equations. Nevertheless, these tasks can be carried to the most difficult problems of mechanics. At the solution of such tasks use the rule of addition of classical speeds. To understand the principle of the decision, it is better to consider it on concrete examples of tasks.
1. Example on the rule of addition of speeds. Let speed currents of the river v0, and speed of the boat crossing this river concerning water be equal to v1 and is directed perpendicular to the coast (cm figure 1). The boat at the same time participates in two independent movements: it for some time of t crosses the river width of N with v1 speed concerning water and for the same time bears her downstream the river on l distance. As a result the boat floats a way of S with v speed concerning the coast, equal on the module: v equally in a root in v1, kvardratny from expression, in a square + v0 in a square for same is a high time for t. Therefore it is possible to write down the equations which solve similar problems: H=v1t, l = v0t? S = root square of expression: v1 in a square + v0 in a square increased by t.
2. Other type of such tasks asks questions: under what corner to the coast to the dolena do to row the oarsman in the boat to appear on the opposite coast, having passed the minimum way during a crossing? For what time will this way be passed? With what speed will the boat pass this way? To answer these questions it is necessary to make the drawing (cm rice 2). It is obvious that the minimum way which there can pass the boat, crossing the river, is equal to width of the river of N. To float this way, the oarsman has to direct the boat under such corner and to a breg, at kotorm the vector of absolute speed of boat v will be directed perpendicular to the coast. Then from a rectangular triangle it is possible to find: cos a=v0/v1. From here it is possible to take a corner and. To determine speed from the same triangle by Pythagorean theorem: v = root square of expression: v1 in a square - v0 in a square. And at last time of t for which the boat will cross the river width of N, moving with v speed, there will be t=H/v.