Question

An ant runs from an ant-hill in a straight line so that its velocity is inversely proportional to the distance from the centre of the ant-hill. When the ant is at point A at a distance l₁ = 1 m from the centre of the ant-hill, its velocity v₁ = 2 cm/s.

What time will it take the ant to run from point A to point B which is at a distance l₂ = 2 m from the centre of the ant-hill?

Answer 1

The velocity of the ant varies with time according to a nonlinear law. Therefore, the mean velocity on different segments of the path will not be the same, and the well-known formulas for mean velocity cannot be used here.

Let us divide the path of the ant from point A to point B into small segments traversed in equal time intervals Δt. Then Δt = Δl/v_m (Δl), where v_m (Δl) is the mean velocity over a given segment Δl. This formula suggests the idea of the solution of the problem: we plot the dependence of 1/v_m (Δl) on 1 for the path between points A and B. The graph is a segment of a straight line (Fig. 134).

The hatched area S under this segment is numerically equal to the sought time. Let us calculate this area:

since 1/v₂ = (1/v₁) l₂/l₁. Thus, the ant reaches point B in the time