A street car moves rectilinearly from station A to the next station B (from rest to rest) with an acceleration varying according to the law `f = a-bx,` where a and b are constants and x is the distance from station A. The distance between the two stations and the maximum velocity are

Options

A. `x=2a/b,v_(max)=a/(sqrtb)`
B. `x=a/2b,v_(max)=a/b`
C. `x=a/2b,v_(max)=b/(sqrta)`
D. `x=a/b,v_(max)=(sqrta)/b`

Answer

A

Explanation

`f=v.(dv)/(dt)=a-bx`
or `int_0^v vdv=int_0^x(a-bx)dx`
`:. v=sqrt(2ax-bx^2)…..(i)`
At other station, `v=0`
`rArr x=(2a)/b`
Further acceleration will change its direction when,
`f=0 or a-bx=0 or x=a/b`
At this x, velocity is maximum.
Using Eq. (i),
`v_(max)=sqrt(2a(a/b)-b(a/b)^2)=a/sqrtb`