Question

a bee is at position 2i+4j+3k at t=0 it moves with a constant velocity v​

Answer 1

Answer:

1

Explanation:

Initial velocity of bee ,  $\vec{u} = (3i +4j) m/s$

Initial position of bee, $\vec{r_i} = (2i+4j+3k)m$

acceleration of bee, $\vec{a} = 0$

At t=2s,

Displacement of particle is given by ,

$\vec{S}=\vec{u}t+0\\ \\ \vec{r_f}-\vec{r_i}=\vec{u}*t\\ \\ \vec{r_f} =\vec{r_i}+\vec{u}t \\ \\ \vec{r_f} = (2i+4j+3k)m/s+(3i+4j)*2\\ \\=>8i+12j+3k$

Hence, final position of bee is given by

$\boxed{(8i+12j+3k)m}$