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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 length $s equals t to the power of 3 end exponent plus 5 comma$ where $s$ is in metre and t is in second. The radius of the path is 20 m. The acceleration of P when t =2s is nearly

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

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

$G i v e n comma blank s equals t to the power of 3 end exponent plus 5$
$S p e e d comma blank v equals fraction numerator d s over denominator d t end fraction equals 3 t to the power of 2 end exponent$
$a n d blank r a t e blank o f blank c h a n g e blank o f blank s p e e d comma blank a subscript t end subscript equals fraction numerator d v over denominator d t end fraction equals 6 t$
$therefore T a n g e n t i a l blank a c c e l e r a t i o n blank a t blank t equals 2 blank s comma$
$a subscript t end subscript equals 6 cross times 2 equals 12 blank m s to the power of negative 2 end exponent$
$a n d blank a t blank t equals 2 s comma blank v equals 3 left parenthesis 2 right parenthesis to the power of 2 end exponent equals 12 m s to the power of negative 1 end exponent$
$therefore C e n t r i p e t a l blank a c c e l e r a t i o n comma blank a subscript c end subscript equals fraction numerator v to the power of 2 end exponent over denominator R end fraction equals fraction numerator 144 over denominator 20 end fraction m s to the power of negative 2 end exponent$
$therefore N e t blank a c c e l e r a t i o n equals a subscript t end subscript superscript 2 end superscript plus a subscript i end subscript superscript 2 end superscript almost equal to 14 m s to the power of negative 2 end exponent$

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