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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1.5 sec

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 sec

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Not Possible

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $1 sec$
Given, $\to r = (t^{2} - 4 t + 6)^i + t^{2}^j$
$\Rightarrow \to v = \frac{d \to r}{d t} = (2 t - 4)^i + 2 t^j$
Also, $\Rightarrow \to a = \frac{d \to v}{d t} = 2^i + 2^j$
As the two vectors i.e. $\to r$ and $\to a$ will be perpendicular to each other
$\Rightarrow \to v \cdot \to a = 0$
$\Rightarrow [(2 t - 4)^i + 2 t^j] \cdot [2^i + 2^j] = 0$
$\Rightarrow 2 (2 t - 4) + 4 t = 0$
$\Rightarrow t = 1 sec$

Suggest Corrections

Similar questions

\to r = (t^{2} - 4 t + 6)^i + (t^{2})^j .

\to r = (3^i - 4^j) m, \to v = (4^i - a^j) m s^{- 1} a n d \to a = (- 6^i + b^j) m s^{- 2}

\to r = a [cos ω t^i + sin ω t^j]

\to r = A (cos t + t sin t)^i + A (sin t - t cos t)^j

t

Join BYJU'S Learning Program