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 P R is drawn normal to the curve at P. Find the instantaneous acceleration of the particle at point P.

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 P R is drawn normal to the curve at P. Find the instantaneous acceleration of the particle at point P. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Slope of line Q R= tan α=-(P Q/Q R) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/960cfa2e0b361980311c58d348b59dd6-.png" /> ⇒ tan α=(-v/1) Slope of line A B at P=(d v/d s)[a t P]=-(1/ tan α) ⇒ (d v/d s)=(1/v) ap=(v d v/d s) =(v × 1/v)=1 m / s 2

Q. 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 $P$ .

Slope of line $QR = tan α = - \frac{PQ}{QR}$

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

Q. 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.

Slope of line QR=tanα=−QRPQ​ ⇒tanα=1−v​ Slope of line AB at P=dsdv​[atP]=−tanα1​ ⇒dsdv​=v1​ap​=dsvdv​ =vv×1​=1m/s2

Q. 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 $P$ .

Slope of line $QR = tan α = - \frac{PQ}{QR}$

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