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If a particle is moving on a curved 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 $(\hat{i} + 2 \hat{j} + 2 \hat{k}) m s^{- 1}$ . Choose the correct option

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

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

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

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

answer is A, C.

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Detailed Solution

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

$∴ \cos θ = \frac{\vec{a} . \vec{v}}{| a | | v |}$
here $θ =$ angle between acceleration a and velocity v

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

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

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

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