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

Answer

2

Explanation

`f = a - bx`
For maximum velocity, acceleration should be zero.
i.e. `a - bx = 0 rArr x = (a)/(b)`
`:.` At `x = (a)/(b)`, the particle has its maximum velocity
`f = (vdv)/(dx) = a - bx rArr (v^(2))/(2) = ax - (bx^(2))/(2) + c`
At `x = 0 , v = 0 rArr c = 0`
Substituting : `x = (a)/(b) ,` gives
`v_(max) = (a)/(sqrt(b))`
Also, the velocity of the car should become zero at station `B`.
i.e. `ax - (bx^(2))/(2) = 0 rArr x = 0 , x = ((2a)/(b))`
`:.` Distance between the two stations is `(2a)(b)`
`:. N = 2`
Alternate : `f = a - bx` means particle will do `SHM`.
At mean position , `f = 0`
`rArr x = (a)/(b) rArr`
In the figure shown, `'C'` is the mean positions and `A` & `B` are extreme positions
`:. x_(max) = (2a)/(b)`