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JEE Main 2007 (AIEEE) · Velocity, Acceleration & Variable Motion

Velocity, Acceleration & Variable Motion question from JEE Main 2007 (AIEEE)

Q39JEE Main 2007 · AIEEEEasy

Velocity, Acceleration & Variable Motion

The velocity of a particle is v = v0 + gt + ft2. If its position is x = 0 at t = 0, then its displacement after unit time (t = 1) is

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Answer: (c)

  1. x=∫01(v0+gt+ft2) dtx = \int_0^1 (v_0 + gt + ft^2)\,dt with x(0)=0x(0) = 0
  2. x=v0t+gt22+ft33x = v_0 t + \dfrac{g t^2}{2} + \dfrac{f t^3}{3} at t=1t = 1
  3. x=v0+g2+f3x = v_0 + \dfrac{g}{2} + \dfrac{f}{3}
  4. So the answer is (c).
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