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 centripetal $5 m / s^{2}$ 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:

14

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\sqrt{80}

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10

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12

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Solution

The correct option is B $10$
According to the question, we can draw the following diagram.
From the diagram,
$a_{p} = 5^i + 8^j$
and $a_{Q} = -^i$
Now, $a_{r e l} = a_{P} - a_{Q}$
$a_{r e l} = 5^i + 8^j - (-^i)$
$a_{r e l} = 6^i + 8^j$
$| a_{r e l} | = \sqrt{(6)^{2} + (8)^{2}} = \sqrt{36 + 64}$
$= \sqrt{100}$
$= 10 m / s^{2}$ .

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The correct option is B 10According to the question, we can draw the following diagram.From the diagram,ap=5^i+8^jand aQ=−^iNow, arel=aP−aQarel=5^i+8^j−(−^i)arel=6^i+8^j|arel|=√(6)2+(8)2=√36+64=√100=10 m/s2.