First slide

Relative motion in 2 dimension

Question

Velocity versus displacement curve of a particle moving in straight line is shown in the figure. From a point P, d 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.

Moderate

Solution

$Slope of line Q R = \tan α = - \frac{P Q}{Q R}$
$\Rightarrow \tan α = \frac{- v}{1}$
$Slope of line A B at P = \frac{d v}{d s} [a t P] = - \frac{1}{\tan α}$
$\Rightarrow \frac{d v}{d s} = \frac{1}{v} a_{p} = \frac{v d v}{d s} = \frac{v \times 1}{v} = 1 m / s^{2}$

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