Question

Two particles P and Q are moving on a circle. At a certain instant of time both the particles are diametrically opposite and P has tangential acceleration $8 \, m/s^2$ and centripetal acceleration $5 \, m/s^2$ whereas Q has only centripetal acceleration of $1 \, m/s^2$. At that instant acceleration (in $m/s^2)$ of P with respect to Q is :

• A. 12
• B. 14
• C. $\sqrt{80}$
• D. 10

Answer 1

The Correct Option is D

According to the question, we can draw the following diagram

From the diagram
$ a _{p} =5 \hat{ i }+8 \hat{ j } $
and $a _{Q} =-\hat{ i }$
Now, $a _{\text {rel }} = a _{p}- a _{Q} $
$ a _{\text {rel }} =5 \hat{ i }+8 \hat{ j }-(-\hat{ i })$
$ a _{\text {rel }} =6 \hat{ i }+8 \hat{ j }$
$\left| a _{\text {rel }}\right| =\sqrt{(6)^{2}+(8)^{2}}=\sqrt{36+64}$
$=\sqrt{100} $
$=10 \,m / s ^{2} $