Q98 / 180 · NEET 2014 · Physics

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. The average velocity vector V→av from t=0 to t=5 s is:

  • A
    1513i^+14j^
  • B
    73i^+j^
  • C
    2i^+j^
  • Correct answer
    D
    115i^+j^

Solution

The position vector of the particle at time t=0 is r→i=2i^+3j^
The position vector of the particle at time t=5 s is r→f=13i^+14j^
The displacement of the particle is s→=11i^+11j^
The average velocity vector V→=11i^+11j^5

Hence, the correct answer is option: 'D'.

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