Ants can make you go crazy!

Classical Mechanics Level 2

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. Then find the time taken by the ant to run from A to B.

NOTE:Please give your answer in seconds.


The answer is 75.

Sumukh Bansal by Sumukh Bansal , 16, India

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1 solution

Steven Chase
Oct 23, 2017

d x d t = α x \frac{dx}{dt} = \frac{\alpha}{x}

Apply the point A condition:

0.02 = α 1 α = 0.02 0.02 = \frac{\alpha}{1} \\ \alpha = 0.02

Start with the first relationship and cross-multiply:

x d x = α d t d t = 50 x d x x \, dx = \alpha \, dt \\ dt = 50 x \, dx

Determine the time from A to B:

Δ t A B = 50 ∫ 1 2 x d x = 50 ( 4 2 − 1 2 ) = 75 \Delta t_{AB} = 50 \int_1^2 x \, dx = 50\big(\frac{4}{2} - \frac{1}{2}\big) = \boxed{75}

3 Helpful 0 Interesting 1 Brilliant 0 Confused

From the given data if v be the velocity at a distance x then

v=k/x

but at x=1 v=2 so k=2 hence

v=2/x but v=dx/dt

so dx/dt=2/x

or xdx=2dt integrating both sides

x^2/2=2t+C c=const of int.

Rearranging we get

t=x^2/4-C/2

From the given data let at t1 x=1

t1=1/4-C/2

and at B x=2 henc

t2=4/4-C/2=1-C/2

So the the time taken from A to B is

t2-t1=1-1/4=3/4 sec=75sec

KARTHIK SHAMSUNDAR - 9 months, 3 weeks ago

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