The velocity of a particle is $v=v_{0}+g t+\mathrm{Ft}^{2}$. Its position is $\mathrm{x}=0$ at $\mathrm{t}=0$; then its displacement after time ($\mathrm{t}=1$) is:

(A) $v_{0}+\frac{g}{2}+\frac{F}{3}$

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Kinematics

The velocity of a particle is

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

. Its position is

x = 0

t = 0

; then its displacement after time (

t = 1

) is:

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

Answer: (A)

Solution

Displacement is the time integral of the velocity:

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}

Putting

t = 1

gives a displacement of

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

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Single correct option

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