Q.

A particle is moving such that its position coordinates (x, y) are (2 m, 3 m) at time t = 0, (6 m, 7 m) at time t = 2 s and (13 m, 14 m) at time t = 5 s. Average velocity vector (vav)from t = 0 to t = 5 s is               [2014]
 

1 15(13i^+14j^)  
2 73(i^+j^)  
3 2(i^+j^)  
4 115(i^+j^)  

Ans.

(4)

At time t = 0, the position vector of the particle is r1=2i^+3j^.

At time t = 5 s, the position vector of the particle is r2=13i^+14j^.

Displacement from r1 to r2 is

Δr=r2-r1=(13i^+14j^)-(2i^+3j^)=11i^+11j^

  Average velocity,  vav=ΔrΔt=11i^+11j^5-0=115(i^+j^)