An ant runs from an anthill in a straight line so that its velocity is inversely proportional to the distance from the center of the anthill. When the ant is at point A at a distance 1 m from the center of the anthill, its velocity is 2 cm/s. What time will it take the ant to run from point A to point B which is at a distance 2m from the center of the anthill?

Solution.

First of all, we can write kinematics equation

v = k/x

where v – velocity of the ant, x – position of the ant from the center of the anthill and k – some constant.

We know that at point A x = 1 m or 100 cm and v = 2 cm/s.

Hence,

2 = k/100 or k = 200 cm^2/s.

Now, the kinematics equation will be

v = 200/x

We should rewrite this equation

1/v = x/200 or

** 1/v = 0.005*x **

and draw a graph of this equation.

Now, we should use the fact that **time is represented by the area** between the function and the x-axis.

In our case, it will be the area of the trapezium.

The area of the trapezium is given by the following formula

h*(a+b)/2

where a and b are the lengths of the parallel sides and h is the perpendicular distance between the parallel sides.

Finally,

100*(0.5+1)/2 = 75 s.

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