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ant runs from an ant hill in a straight line so that its velocity is inversely proportional to the distance from the centre of ant hill. when the ant is at point A at a distance of 1 m from the centre of the hill its velocity is 2 centimetre per second.point B is at a distance of 2 m from the centre of the ant hill. the taken by the ant to run from a to b is

Krishna patil , 5 Years ago

Grade 9

2 Answers

Arun

Last Activity: 5 Years ago

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

Parmendra Kumar Bajpai

Last Activity: 5 Years ago

In the previous answer, their is dimensional error.

From the given data if v be the velocity at a distance x then
v=k/x
but at x=100 cm, v=2 cm/sec so k=200 cm2/sec. Now

from the definition, v=dx/dt,
therefore, dx/dt = 200/x
or xdx=200dt integrating both sides
x^2/2=200t+C, where C is the constant of integration.
Rearranging we get
t=x^2/400-C/200
From the given data let at t1 x=100 cm
t1=100x100/400-C/200
and at B x=200 cm hence
t2=200x200/400-C/200
So the the time taken from A to B is
t2-t1=100-25 =75sec

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