Position vector of a particle is given as vec r, = ,({t^2}, - ,8t, + ,12),hat{i},, + ,,{t^2}hat{j} t

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3-2.Motion in Plane

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The position vector of a particle is given as $\vec r\, = \,({t^2}\, - \,8t\, + \,12)\,\hat i\,\, + \,\,{t^2}\hat j$ The time after which velocity vector and acceleration vector becomes perpendicular to each other is equal to........$sec$

A

$1$

B

$2.5$

C

$2$

D

$1.5$

Solution

$\mathrm{V}=\frac{\mathrm{dx}}{\mathrm{dt}}=(2 \mathrm{t}-8) \hat{\mathrm{i}}+2 \mathrm{t\hat j}$

$a=\frac{d v}{d t}=2 \hat{i}+2 \hat{j}$

$\overrightarrow{\mathrm{V}} \cdot \overrightarrow{\mathrm{a}}=0$

$4 t-16+4 t=0$

$t=2 \mathrm{sec}$

Standard 11

Physics

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