Question

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 100cm from the center
of the hill, its velocity is 2 cm/s. Point B is at a
distance of 200 cm from the center of the ant-hill. The
time taken by the ant to run from A to B is​

Answer 1

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

Mark me brainest Frist

Answer 2

Answer:

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

Mark me brainliest plss