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Grade: 11
        
The initial velocity of a particle moving along x axis is u (at t=0 and x=0) and its acceleration is given by a=kx .which of the reLation is correct.                                                                                                           V^2-u^2=2kx.    Or.        V^2=u^2+ kx^2
4 years ago

Answers : (1)

Rinki jain
12 Points
							a=kxa=vdv/dxadx=vdvkxdx=vdvkx^2/2=v^2/2kx^2/2+C = v^2/2if x=0 , velocity = u C = u^2/2thus,v^2/2 = kx^2/2 + u^2/2 v^2 = u^2 + kx^2we have a = kxv^2 = u^2 + 2ax       = u^2+ 2kx.xv^2 = u^2 + 2kx^2
						
2 years ago
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