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

1 sec

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2 sec

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1.5 sec

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Solution

The correct option is A 1 sec
$\to r = (t^{2} - 4 t + 6)^i + (t^{2})^j$
velocity $= \frac{d r}{d t} = (2 t - 4)^i + (2 t)^j$
acceleration $= \frac{d v}{d t} = (2)^i + (2)^j$
$\to a . \to v = ((2 t - 4)^i + (2 t)^j) . ((2)^i + (2)^j)$
when they are perpendicular then a.v = 0
$((2 t - 4)^i + (2 t)^j) . ((2)^i + (2)^j) = 0$
$(2 t - 4) \times 2 + 4 t = 0$
$4 t - 8 + 4 t = 0$
$t = 1 s$

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

The correct option is A 1 sec→r=(t2−4t+6)^i+(t2)^jvelocity =drdt=(2t−4)^i+(2t)^jacceleration =dvdt=(2)^i+(2)^j→a.→v=((2t−4)^i+(2t)^j).((2)^i+(2)^j)when they are perpendicular then a.v = 0 ((2t−4)^i+(2t)^j).((2)^i+(2)^j)=0(2t−4)×2+4t=04t−8+4t=0t=1s

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