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. Find the instantaneous acceleration of the particle at point P

__

Answer 1

Acceleration = $\overrightarrow{a}$ = $\frac{dv}{dt}=\frac{ds}{dt}\frac{dv}{ds}=\frac{vdv}{ds}$

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 $\frac{y-0}{2-3}$ = -y

Sp slope of tangent which will be perpendicualr to normal will be $\frac{1}{y}$

So that product = -1

-y * $\frac{1}{y}$ = -1

So $\frac{dv}{ds}=\frac{1}{y}$ (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 )

$\Rightarrow \overrightarrow{a}=y\times \frac{1}{y}=1$

a = 1m/$s^2$