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

Answer

A

Explanation

For maximum velocity, acceleration should be zero.
i.e., `a-bx=0impliesx=(a)/(b)`
At `x=(a)/(b)`, the particle has its maximum velocity
`f=(vdv)/(dx)=a-bximplies(v^2)/(2)=ax-(bx^2)/(2)+c`
At `x=0`, `v=0impliesc=0`
`v_(max)=(a)/(sqrtb)`
Aslo, the velocity of the car should become zero at station B.
i.e., `ax-(bx^2)/(2)=0impliesx=0`,`x=((2a)/(b))`
Therefore ditance between the two stations is `(2a)/(b)`. Alternate:`f=a-bx` means particle will do `SHM`. At mean position `f=0` `implies=(a)/(b)`
In the figure shown C is the mean position and A and B are extreme position Thus,
`x_(max)=(2a)/(b)` and `V_(max)=omegaA=sqrtb(a)/(b)=(a)/(sqrtb)`