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

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

1.5

1 s

5 s

2 s

answer is B.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\vec{r} = (t^{2} - 4 t + 6) \hat{i} + t^{2} \hat{j}; \vec{v} = \frac{\vec{dr}}{dt}$

$= (2 t - 4) \hat{i} + 2 t \hat{j}, \vec{a} = \vec{\frac{dv}{dt}} = 2 \hat{i} + 2 \hat{j}$

$If \vec{a} and \vec{v} are perpendicular, \vec{a} . \vec{v} = 0$

$(2 \hat{i} + 2 \hat{j}) . ((2 t - 4) \hat{i} + 2 t \hat{j}) = 0$

8t-8 = 0, t = 1 sec.

Watch 3-min video & get full concept clarity

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test