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A particle is moving such that its position coordinates (x,y) are (2m, 3m) at time t = 2s and (13 m, 14m) at time t = 5s
Average velocity vector from t = 0 to t = 5 s is
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Correct option is D
At time t = 0, the position vector of the particle is r→1=2i^+3j^At time t = 5s, the position vector of the particle is r→2=13i^+14j^Displacement from r→1 to r→2 isΔr→=r→2−r→1=(13i^+14j^)−(2i^+3j^)=113i^+11j^∴Average velocity, v→av=Δr→Δt=11i^+11j^5−0=115(i^+j^)Talk to our academic expert!
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