The position vector of a particle is given as | vedclass

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Question

$\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{v}}=0$
$(2 \hat{i}+2 \hat{j}) \cdot((2 t-4) \hat{i}+2 t \hat{j})=0$
$8 t-8=0$  &nbsp

Vedclass · Accepted Answer

$1$

Vedclass · Answer

$\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{v}}=0$
$(2 \hat{i}+2 \hat{j}) \cdot((2 t-4) \hat{i}+2 t \hat{j})=0$
$8 t-8=0$              $ t=1 \sec$

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

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