Physics - Kinematics Question with Solution | TestHub

PhysicsKinematics1-DMedium2 minPYQ_2021

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

Options:

Answer:

Solution:

$v = v_{0} + g t + F t^{2}$

$\frac{d s}{d t} = v_{0} + g t + F t^{2}$

$\int d s = \int_{0}^{1} (v_{0} + g t + F t^{2}) d t$

$s = {[v_{0} t + \frac{g t^{2}}{2} + \frac{F t^{3}}{3}]}_{0}^{1}$

$s = v_{0} + \frac{g}{2} + \frac{F}{3}$

Stream:JEESubject:PhysicsTopic:KinematicsSubtopic:1-D

⏱ 2mℹ️ Source: PYQ_2021

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