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Question

Velocity versus displacement curve of a particle moving in straight line is shown in the figure. From a point P, a line is drawn perpendicular to displacement axis and line PR is drawn normal to the curve at P. The instantaneous acceleration of the particle at point P___.

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Solution

Acceleration = a = dvdt=dsdtdvds=vdvds

Coordinates of point p = (2,y)

( Since P and Q lie on a line perpendicular to x axis )

Now we know PR is a line

Normale to curve

Its slope is y023 = -y

Sp slope of tangent which will be perpendicualr to normal will be 1y

So that product = -1

-y * 1y = -1

So dvds=1y (since dv/ds gives the slope of the curve, and at that point, we got the slope as 1/y)

Now velocity at point p = y ( from the graph , y axis is velocity )

a=y×1y=1

a = 1m/s2


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