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The position vector of a particle is given as $\vec{r} = (t^{2} - 4 t + 6) \hat{i} + (t^{2}) \hat{j}$ . Find the time after which the velocity vector and acceleration vector becomes perpendicular to each other.

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Correct option is B

r→ = (t2−4t+6)i^+t2j^; v→ = dr→dt= (2t−4)i^+2t j^, a→= dvdt→ = 2i^+2j^If a→ and v→ are perpendicular, a→.v→ = 0(2i^+2j^).((2t−4)i^+2t j^)=08t-8 = 0, t = 1 sec.

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The position vector of a particle is given as r→= (t2−4t+6)i^+(t2)j^. Find the time after which the velocity vector and acceleration vector becomes perpendicular to each other.

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Ready to Test Your Skills?

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The position vector of a particle is given as $\vec{r} = (t^{2} - 4 t + 6) \hat{i} + (t^{2}) \hat{j}$ . Find the time after which the velocity vector and acceleration vector becomes perpendicular to each other.