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The velocity v of a particle moving along x axis varies with its position x as shown in figure. The acceleration of a particle varies with position as
Dear student,Acceleration of a particle=Velocity as a function of displacement can be writtten from the given figure as-Hence
a==dv/dtdv/dx×dx/dta=vdv/dxFrom graph dv/dx=2a=-2v........1- indicates line of graphNow using eqn of slopey=mx+cv=-2+4- is due to negative slope a=-2(-2x+4)a=4x-8
Dear student,Please find the solution to your problem. a = dv/dtdv/dx × dx/dta = vdv/dxFrom graph dv/dx = 2a = – 2v ........1 – sign indicates line of graphNow using eqn of slopey = mx + cv = – 2x + 4 – sign is due to negative slope a = – 2(–2x + 4)a = 4x – 8 Thanks and regards,Kushagra
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