Question

# Derive an equation for position-velocity relation (2as=v2−u2) by graphical method

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Solution

## Let the initial velocity of the object = uLet the object is moving with uniform acceleration, a.Let object reaches at point B after time, t and its final velocity becomes, vDraw a line parallel to x-axis DA from point, D from where object starts moving.Draw another line BA from point B parallel to y-axis which meets at E at y-axis.The distance covered by the object moving with uniform acceleration is given by the area of trapezium ABDOTherefore, Area of trapezium ABDOE =12× (sum of parallel sides + distance between parallel sides)Distance(S)=12(DO+BE)×OES=12(u+v)×t...............(i)we know that,a=v−ut from above equation we can say,t=v−ua...........(ii)After substituting the value of t from equation(ii) in equation (i)S=12a(u+v)(v−u)2aS=(u+v)(v−u)2aS=v2−u2Hence Proved.

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