The position vector of a particle is given as r^> = (t^2-4t+6)i^ + (t^2)j^. What is the time after which the velocity vector and acceleration vector becomes perpendicular to each other?

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Solution

$\vec{r} = (t^{2} - 4 t + 6) \overset{⏜}{i} + (t^{2}) \overset{⏜}{j} v e l o c i t y = \frac{d r}{d t} = (2 t - 4) \overset{⏜}{i} + (2 t) \overset{⏜}{j} a c c e l e r a t i o n = \frac{d v}{d t} = (2) \overset{⏜}{i} + (2) \overset{⏜}{j} \vec{a} . \vec{v} = ((2 t - 4) \overset{⏜}{i} + (2 t) \overset{⏜}{j}) . ((2) \overset{⏜}{i} + (2) \overset{⏜}{j}) w h e n t h e y a r e p e p e n d i c u l a r t h e n a . v = 0 ((2 t - 4) \overset{⏜}{i} + (2 t) \overset{⏜}{j}) . ((2) \overset{⏜}{i} + (2) \overset{⏜}{j}) = 0 (2 t - 4) \times 2 + 4 t = 0 4 t - 8 + 4 t = 0 t = 1 s$

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