A particle retards from a velocity `v_(0)`while moving in a straight line. If the magnitude od deceleration is directly proportional to thesquare roop of the speed of the particle, find its average velocity for thetotal tome of ist motion.

Options

A. `(v_(0))/(3)`
B. `(3v_(0))/(2)`
C. `(2v_(0))/(3)`
D. `(2v_(0))/(5)`

Answer

C

Explanation

The average velocity is given as:
`v_(av)=(S)/(t)=(Total displacment)/(Total time)`
We know `a=v(dv)/(ds) rArr v dv =a ds`
Let us calculate the displacement by substituting `a=-alpha sqrt v`
(for retardation) in `v dv =a xds`.
We have `v dv =-alpha sqrt v` ds. `rArr sqrt v ds =- alpha ds`
When the particle slows sown from `v_(0)` to ` 0`, it comers a distance `s`.
Hence, `int_(v0)^(0) sqrt dv =- int _(0)^(s) alpha` ds
This gives `s =(2v_(0)^(3//2))/(3 alpha)`
Use of `a=(dv)/(dt)`: Now we will find the time of motion by substituting `a =- sqrtv`
in `a(dv)/(dt)`, we get
`-alpha sqrtv=(dv)/(dt) rArr (dv)/(sqrtv) =- alpha dt`
If the particle takes time `t` to stop, we have
`int_(v0)^(0) (dv)/(sqrtv) =-alpha int_(0)^(t) dt`
This yields `t=(sqrtv_(0))/(alpha)`.
Finally sunstituting `s` and (t) in `v_(av)=(s)/(t)`, we have
`v_(av)=(2v_(0))/(3)`.