Position vector of a particle is given as vec r = (t^2 - 4t + 6)hat{i} + (t^2 )hat{j} . time after w

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3-2.Motion in Plane

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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.......$sec$

A$1$

B$2$

C$1.5$

DNot possible

Solution

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

Standard 11

Physics

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