If a particle starts from A along the curved circular path shown in figure with tangential acceleration ‘a’. Then acceleration at B in magnitude is:

$a π \sqrt{1 + π^{2}}$

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$a \sqrt{π^{2} - 1}$

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$a \sqrt{1 + π^{2}}$

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$2 a \sqrt{1 + π^{2}}$

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Solution

The correct option is C
$a \sqrt{1 + π^{2}}$

$V_{B}^{2} = V_{0}^{2} + 2 a \times \frac{π R}{2}$

$V_{B}^{2} = a π R$

Normal acceleration at $B = \frac{V_{B}^{2}}{R} = a π$

Net acceleration at $B = \sqrt{a^{2} + a^{2} π^{2}} = a \sqrt{1 + π^{2}}$

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If a particle starts from A along the curved circular path shown in figure with tangential acceleration ‘a’. Then acceleration at B in magnitude is:

The correct option is C a√1+π2 V2B=V20+2a×πR2 V2B=aπR Normal acceleration at B=V2BR=aπ Net acceleration at B=√a2+a2π2=a√1+π2

The correct option is C
$a \sqrt{1 + π^{2}}$

$V_{B}^{2} = V_{0}^{2} + 2 a \times \frac{π R}{2}$

$V_{B}^{2} = a π R$

Normal acceleration at $B = \frac{V_{B}^{2}}{R} = a π$

Net acceleration at $B = \sqrt{a^{2} + a^{2} π^{2}} = a \sqrt{1 + π^{2}}$