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Consider a particle in circular motion as shown in the figure. The position vector sweeps equal area in equal time. Select correct option(s).

(A) the speed is constant
(B) acceleration is constant
(C) $\left| {\overrightarrow {{a_r}} } \right|$ = constant
(D) $a_t = 0$

Let $d\theta$ be the angle swept by the radius in time dt.

Area swept $dA = \frac{1}{2}{r^2}d\theta$

$\frac{{dA}}{{dt}} = \frac{1}{2}{r^2}\frac{{d\theta }}{{dt}} = \frac{1}{2}{r^2}\omega$ where $\omega$ is the instantaneous angular speed.

$\frac{1}{2}{r^2}\omega$ = constant

$\therefore \omega$ = constant or speed v = constant

If speed is constant, $a_t = 0$

${a_r} = \frac{{{v^2}}}{r}$ = constant (magnitude)

The direction of acceleration vector is changing and hence it is not a constant.

(A), (C), (D).

### A man starts walking from the point P (-3, 4) ....

A man starts walking from the point P (-3, 4), touches the x-axis at R, and then turns to reach at the point Q (0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $50 [(PR)^2 + (RQ)^2 ]$ is equal to _ _ _ _ . Solution For time to be minimum at constant speed, the directions must be symmetric. In other words, the angles made by PR and RQ with the vertical must be the same just like in the law of reflection in optics. $tan \theta = \frac {MP}{MR} = \frac {NQ}{NR}$ $\Rightarrow \frac {3-r}{4} = \frac {r}{2}$ $\Rightarrow r=1$ So, $R \equiv ( - 1,0)$ Now, $50(PR^2+RQ^2)=50[(4+16)+(1+4)]=1250$