A particle is moving in a straight line such that `dv//dt=-lambdav`, where `v` is the velocity and `lambda` is a constant. At `t=0`, the particle is at the origin and moves with velocity `v_(0)`. Sketch the following graphs:
`(a) v-t (b) s-t (c ) v-s (d) log_(e)v v//s t`

Explanation

as solved earlier
(a) `v=v_(0)e^(-lambdat)`
Shape: `y=v_(0)e^(-lambdax)`, graph will be of the type `y=e^(-x)`

(b) `s=v_(0)/lambda(1-e^(-lambdat))`
Shape: `y=v_(0)/lambda(1-e^(-lambdat))`, graph will be of the type
`y=1-e^(-x)`

(c ) `v=v_(0)-lambdas`
Shape: `y=v_(0)-lambdax`(straight line)

(d) `v=v_(0)e^(-lambdat)`
`log_(e) v=log_(e)v_(0)-lambdat`
Shape: `y=log_(e)v_(0)-lambdax` (straight line)

(Graph involving log is always a straight line.)