Question

For the two particles $P$ and $Q$ along the diametrically opposite end of a circle $(r=4\text{m})$ as shown in figure at time $t=0$, origin is taken at centre of the circle.

If Particle $P$ moves $6\text{m}$ towards $Q$, along line joining them. Then position of $P$ with respect to $Q$ will be:

• A. $1\text{m}$
• B. $+3\text{m}$
• C. $-6\text{m}$
• D. $-2\text{m}$

Answer 1

The correct option is D $-2\text{m}$

Initially particle $P$ is at $X=-4\text{m}$ from origin $\left(O\right)$.
After moving through $6\text{m}$ along line $PQ$, towards $Q$, it will be at a distance of $2\text{m}$ from origin along $+ve-X$ axis.
$\Rightarrow$ Position of particle $P$ w.r.t origin is,
$X_P=+2\text{m}$
Now position of particle $Q$ w.r.t origin $O$ is,
$X_Q=+4\text{m}$
Hence, position of particle $P$ w.r.t $Q$ is given as;
$\Rightarrow X_{PQ}=X_P-X_Q$
$\Rightarrow X_{PQ}=2-(+4)=-2\text{m}$