A particle is moving such that its position coordinates . . .

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

a) (1/3)(13i+14j)
b) (7/3)(i+j)
c) 2(i+j)
d) (11/5)(i+j)

Doubt by Trishna

Solution : 

We know, 
At t=0 sec
Coordinates of Particle are (2m, 3m)
Hence, the position vector will be 
r1=2i+3j

At t=2 sec
Coordinates of Particle are (6m, 7m)
Hence, the position vector will be 
r2=6i+7j

At t=5 sec
Coordinates of Particle are (13m, 14m)
Hence, the position vector will be 
r3=13i+14j

But we know, 
Average velocity 
= Total Displacement / Total Time

Vav = [r3-r1]/[t3-t1]
Vav = [(13i+14j)-(2i+3j)]/(5-0)
Vav = [13i+14j-2i-3j]/5
Vav = [11i+11j]/5
Vav = (11/5)(i+j)
Hence, d) (11/5)(i+j) would be the correct option.