InterviewSolution
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InterviewSolution
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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 areA. `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` |
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Answer» Correct Answer - A `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` |
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