An ant runs from an ant - hill in a straight line so that its velocity is inversely proportional to the distance from the center of ant - hill. When the ant is at a point A at a distance 1 m from the center of the hill, its velocity is 2 cm/s. Point B is at a distance of 2 m from the center of the ant - hill. The time taken by the ant to run from A to B is

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25 s

35 s

75 s

55 s

answer is B.

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Detailed Solution

According to given condition
Distance is inversely proportional to velocity
$\begin{array}{l} S \propto \frac{1}{v} v_{1} = 2 cm / s \\ \frac{s_{1}}{s_{2}} = \frac{v_{2}}{v_{1}} \\ \frac{1}{2} = \frac{v_{2}}{2} v_{2} = 1 cm / s \end{array}$
When two Quantities are directly proportional to each other graph among them is straight line
Area of the graph under $\frac{1}{v}$ and gives time

$\begin{array}{l} = \frac{1}{2} (s_{2} - s_{1}) [\frac{1}{v_{2}} - \frac{1}{v_{1}}] + (s_{2} - s_{1}) \frac{1}{v_{1}} \\ t = \frac{1}{2} (2 - 1) [\frac{100}{1} - \frac{100}{2}] + (2 - 1) \frac{100}{2} \\ t = \frac{1}{2} \times 1 [\frac{100}{2}] + 1 \times 50 \\ t = 25 + 50 \\ t = 75 s \end{array}$