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Physics-

General

Easy

Question

A point $P$ moves in counter-clockwise direction on a circular path as shown in the figure. The movement of $´ ´ P ´ ´$ is such that it sweeps out a length $s equals t to the power of 3 end exponent plus 5$ , where $s$ is in metres and $t$ is in seconds. The radius of the path is $20 blank m$ . The acceleration of $´ ´ P ´ ´$ when $t equals 2 s$ is nearly

$14 m / s^{2}$
$13 m / s^{2}$
$12 m / s^{2}$
$7.2 m / s^{2}$

The correct answer is: $14 blank m divided by s to the power of 2 end exponent$

As $S equals t to the power of 3 end exponent plus 5$
$fraction numerator d s over denominator d t end fraction equals 3 t to the power of 2 end exponent equals v$
$therefore a subscript t end subscript equals fraction numerator d v over denominator d t end fraction equals 6 t$
at $t equals 2 blank s e c$
$open vertical bar stack a with rightwards arrow on top close vertical bar equals square root of a subscript c end subscript superscript 2 end superscript plus a subscript t end subscript superscript 2 end superscript end root$
$equals square root of open parentheses fraction numerator v to the power of 2 end exponent over denominator R end fraction close parentheses to the power of 2 end exponent plus a subscript t end subscript superscript 2 end superscript end root equals square root of open parentheses fraction numerator 4 t to the power of 4 end exponent over denominator R end fraction close parentheses to the power of 2 end exponent plus open parentheses fraction numerator d v over denominator d t end fraction close parentheses to the power of 2 end exponent end root$
$equals square root of open parentheses 7.2 close parentheses to the power of 2 end exponent plus 144 end root$
$open vertical bar stack a with rightwards arrow on top close vertical bar equals 14 m divided by s to the power of 2 end exponent$

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