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NEET 2014 (AIPMT) · Motion in a Plane (2-D Kinematics)

Motion in a Plane (2-D Kinematics) question from NEET 2014 (AIPMT)

Q17NEET 2014 · AIPMTEasy

Motion in a Plane (2-D Kinematics)

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 (v⃗av)\left(\vec{v}_{av}\right) from t = 0 to t = 5 s is

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Answer: (d)

  1. v⃗av=Δr⃗Δt=(13−2)i^+(14−3)j^5−0\vec v_{av} = \frac{\Delta\vec r}{\Delta t} = \frac{(13-2)\hat i + (14-3)\hat j}{5 - 0}
  2. =115(i^+j^)= \frac{11}{5}(\hat i + \hat j)
  3. So the answer is (d) 115(i^+j^)\frac{11}{5}(\hat i + \hat j).
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