If a particle is moving on a curve path, the component of acceleration in the direction of velocity is tangential acceleration, while the component of acceleration perpendicular to velocity is normal or centripetal acceleration. A particle is moving on a curved path. Its acceleration and velocity at point P is $(\hat{i} + 2 \hat{j} - 2 \hat{k}) m s^{- 2} and (2 \hat{i} + 2 \hat{j} + \hat{k}) m s^{- 1} .$ choose the correct option (s).

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The tangential acceleration at point P is $\frac{4}{3} m s^{- 2}$

The normal acceleration at point P is $\frac{\sqrt{65}}{3} m s^{- 2}$

The normal acceleration at point P is $\frac{4}{3} m s^{- 2}$

The tangential acceleration at point P is $\frac{\sqrt{65}}{3} m s^{- 2}$

answer is A, D.

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Detailed Solution

$∴ \vec{a} . \vec{v} = | \vec{a} | | \vec{v} | \cos θ, ∴ \cos θ = \frac{a . v}{| a | | v |}$

Here , $θ=angle between acceleration \vec{a} and velocity \vec{v}, ∴ \cos θ = \frac{2 + 4 - 2}{3 \times 3} = \frac{4}{9}$

The magnitude of acceleration is $a = | \vec{a} | = \sqrt{1 + 4 + 4} = 3 m s^{- 2}$

$∴$ The tangential acceleration is $a_{t} = a c o s θ = 3 \times \frac{4}{9} = \frac{4}{3} m s^{- 2}$

And normal acceleration is

$a_{n} = a \sin θ = 3 \sqrt{1 - \cos^{2} θ} = 3 \sqrt{1 - \frac{16}{81}} = \frac{3}{9} \sqrt{65} = \frac{\sqrt{65}}{3} m s^{- 2}$

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