Question

A particle is moving in a straight line such that its velocity varies as `v=v_(0) e^(-lambdat)`, where `lambda` is a constant. Find average velocity during the time interval in which the velocity decreases from `v_(0)` to `v_(0)/2`.

Answer 1

`v=v_(0) e^(-lambdat)`
`t=0, v=v_(0)`
When `v=V_(0)/2, t=t_(0)=?`
`v_(0)/2=v_(0)e^(-lambdat_(0))implies e^(-lambdat_(0))=1/2`
`e_(lambdat_(0))=2`
Taking log on base `e`, we get
`lambdat_(0) log_(e) e=log_(e)2`
`t_(0)=(log_(e)2)/lambda`
Average velocity for time duration `t=0` to `t=t_(0)`,
`bar(v)=(int_(0)^(t_(0)) v dt)/(int_(0)^(t_(0)) dt)=(v_(0)int_(0)^(t_(0)) e^(-lambdat) dt)/(|t|_(0)^(t_(0))=(t_(0)-0))`
`=v_(0)/t_(0)(|e^(-lambdat_(0))-e^(0)|)/-lambda`
`=v_(0)/(lambdat_(0))(1-e^(-lambdat_(0)))`
`=v_(0)/(lambdaT_(0))(1-1/2)`
`=v_(0)/(2lambdat_(0))`
`=v_(0)/(2lambda.(log_(e)2)/lambda)=v_(0)/(2 log_(e)2)`