A body moves in a straight line under the retardation #a# which is given by #a=-bv^2#, (#b# is a positive constant and #v# represents the velocity). If the initial velocity is #2m/s#, what is the distance covered in time #t#=2 seconds?

1 Answer
A08
Apr 9, 2018

Retardation is given by the expression

#a=-bv^2#
#=>(dv)/dt =-bv^2#

Assuming that velocity changes by an amount #dv# in infinitesimal time interval #dt#. Above expression can be rewritten as

#(dv)/v^2 =-b dt#

Integrating both sides with respective variable we get

#int (dv)/v^2 =-int\ b\ dt#
#=>-1/v=-bt+C#
where #C# is a constant of integration. To be ascertained from initial conditions. At #t=0# we have #v=2\ ms^-1#

We get

#-1/2=C#

Expression for velocity becomes

#-1/v=-bt-1/2#
#=>1/v=(1+2bt)/2#
#=>v=2/(1+2bt)#
#=>(dS)/dt=2/(1+2bt)#

Assuming that displacement #dS# takes place in infinitesimal time interval #dt#. Above expression can be rewritten as

#dS=2/(1+2bt)dt#

Integrating both sides with respective variables we get

#int\ dS=int\ 2/(1+2bt)dt#
#=>S=ln(|1+2bt|)/b+C_1#
where #C_1# is a constant of integration.

At #t=0#, #S=0#. Inserting in above we get #C_1=0# and expression for displacement becomes.

#S=ln(|1+2bt|)/b#

Therefore, distance covered in time #t=2\ s#

#S(2)=[ln(1+4b)] /b#

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