The velocity of a particle is $v = v_0 + gt + ft^{2}$. If its position is $x = 0$ at $t = 0$, then its displacement after unit time ($t = 1$) is

(B) $v_0 + \dfrac{g}{2} + \dfrac{f}{3}$

JEE Main · 2007 · 29 Apr 2007 · Offline · Kinematics

The velocity of a particle is

v = v_{0} + g t + f t^{2}

. If its position is

x = 0

t = 0

, then its displacement after unit time (

t = 1

) is

(A) $v_{0} + 2 g + 3 f$
(B) $v_{0} + \frac{g}{2} + \frac{f}{3}$ ✓ Correct
(C) $v_{0} + g + f$
(D) $v_{0} + \frac{g}{2} + f$

Answer: (B)

Solution

Position is the time integral of velocity, so with

x = 0

t = 0

the displacement is

x = \int_{0}^{t} (v_{0} + g t^{'} + f t^{' 2}) d t^{'} = v_{0} t + \frac{g t ^{2}}{2} + \frac{f t ^{3}}{3}

. Evaluating this at

t = 1

gives a displacement of

v_{0} + \frac{g}{2} + \frac{f}{3}

JEE Main · 2007 · 29 Apr 2007 · Offline · Single correct option

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