A particle is moving such that its position is given byx - A cos- ty - A sin- tz - A - tSpeed of particle will beA- - AB- -2- AC- - A -2D- Zero

The correct option is B √(2)ω Ar→ = A cos ω î + A sin ω t ĵ + A ω t k̂ v→ = A ω sin ω t î+ A ω cos ω t ĵ+A ωk̂ |v→|=ω A √(( sin ω t)^2 + (cos ω t)^2 + (1)^2) ...

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Standard XII

Mathematics

Point of Inflection

A particle is...

Question

A particle is moving such that its position is given by

x = A cos $ω t$
y = A sin $ω t$
z = A $ω t$
Speed of particle will be

ω A

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\sqrt{2} ω A

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\frac{ω A}{2}

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Zero

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Solution

The correct option is B $\sqrt{2} ω A$
$\to r = A c o s ω^i + A s i n ω t^j + A ω t^k \to v = - A ω s i n ω t^i + A ω c o s ω t^j + A ω^k | \to v | = ω A \sqrt{(- s i n ω t)^{2} + (c o s ω t)^{2} + (1)^{2}} = \sqrt{2} ω A$

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A particle is moving such that its position is given by x = A cos ωt y = A sin ωt z = A ωt Speed of particle will be

The correct option is B √2ωA→r=A cos ω ^i+A sin ωt^j+A ωt^k→v=−Aω sin ωt ^i+Aω cos ωt^j+Aω^k|→v|=ωA√(−sin ωt)2+(cos ωt)2+(1)2=√2ωA

A particle is moving such that its position is given by

x = A cos $ω t$
y = A sin $ω t$
z = A $ω t$
Speed of particle will be

The correct option is B $\sqrt{2} ω A$
$\to r = A c o s ω^i + A s i n ω t^j + A ω t^k \to v = - A ω s i n ω t^i + A ω c o s ω t^j + A ω^k | \to v | = ω A \sqrt{(- s i n ω t)^{2} + (c o s ω t)^{2} + (1)^{2}} = \sqrt{2} ω A$