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A street car moves rectilinearly from station A to the next station B 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=(ab -1/2 ) b) x=(b/2a), v=(a/b) c) x=(a/2b), v=(ba -1/2 ) d) x=(a/b), v=(a 1/2 /b)


A street car moves rectilinearly from station A to the next station B 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=(ab-1/2)         b) x=(b/2a), v=(a/b)         c) x=(a/2b), v=(ba-1/2)          d) x=(a/b), v=(a1/2/b)


Grade:upto college level

1 Answers

Shivam
29 Points
5 years ago
Acceleration f=a−bxf=a−bx … … … (1)The acceleration is decreasing with increasing x hence velocity will be maximum where f=0f=0. From equation (1) this point is at x=abx=ab.Further, f=dvdt=dvdt.dxdx=vdvdxf=dvdt=dvdt.dxdx=vdvdxWith this we can write equation (1) asvdvdx=a−bxvdvdx=a−bxOr, vdv=(a−bx)dxvdv=(a−bx)dx… … … … (2)For maximum velocity vmaxvmax, we differentiate equation (2) with proper limits.∫vmax0vdv=∫ab0(a−bx)dx∫0vmaxvdv=∫0ab(a−bx)dx|v22|vmax0=|ax−bx22|ab0|v22|0vmax=|ax−bx22|0abOr, vmax=ab√vmax=abFor the distance SS between the two stations, we integrate equation (2) noting that velocity at both the station is zero.∫00vdv=∫S0(a−bx)dx∫00vdv=∫0S(a−bx)dxOr, 0=|ax−bx22|S00=|ax−bx22|0SThis gives S=2ab

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