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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)

$x = \int_0^1 (v_0 + gt + ft^2)\,dt$ with $x(0) = 0$
$x = v_0 t + \dfrac{g t^2}{2} + \dfrac{f t^3}{3}$ at $t = 1$
$x = v_0 + \dfrac{g}{2} + \dfrac{f}{3}$
So the answer is (c).

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