A particle is moving such that its position coordinates `(x, y)` are `(2m, 3m)` at time `t=0, (6m, 7m)` at time `t=2 s`, and `(13 m, 14m)` at time `t=5 s`.
Average velocity vector`(vec(V)_(av))` from `t=0` to `t=5 s` is

Options

A. `(1)/(5)(13veci+14hatj)`
B. `(7)/(3)(hati+hatj)`
C. `2(hati+hatj)`
D. `(11)/(5)(hati+hatj)`

Answer

d

Explanation

Average velocity `hatv_(av)=("total displacement"(vecs))/("total tiime taken(t)")`
Initial position `vecr_(1)=2hati+3hatj`
Final position, `vecr_(2)=13hati+14hatj`
Displacement, `vecs=vecr_(2)-vecr_(1)`
`vecv_(av)=(vecr_(2)-vecr_(1))/(t_(2)-t_(1))`
`vecv_(av)=((13hati+14hatj)-(2hati+3hatj))/(5-0)`
`vecv_(av)=(11hati+11hatj)/(5)`
`vecv_(av)=(11)/(5)(hati+hatj)`