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Question

# A particle moves in the xy-plane with constant acceleration a directed along the negative y-axis. The equation of path of the particle has the form y=bx−cx2, where b and c are positive constants. The velocity v of the particle at the origin of coordinates will be

A
a2c(1+b2)
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B
a4c(1+b2)
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C
3a2c(1b2)
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D
none of these
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Solution

## The correct option is C √a2c(1+b2)x=vxtvy=dydt=bdxdt−2cxdxdt=bvx−2cxvxtherefore vy at x =0 y =0 isvy=bvxS=ut+12at2y=vyt−0.5at2vy=bvx0.5a=cv2xv=√v2x+v2y=√v2x+(bvx)2=√v2x(1+b2)=√a2c(1+b2)

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