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?
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.
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
where a and b are the lengths of the parallel sides and h is the perpendicular distance between the parallel sides.
100*(0.5+1)/2 = 75 s.
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