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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

Option: 1

v_{0}+g/2+f


Option: 2

v_{0}+2g+3f


Option: 3

v_{0}+g/2+f/3


Option: 4

v_{0}+g+f


Answers (1)

best_answer

:     Given :velocity \: \nu = \nu _{0}+gt+ft^{2}

\therefore \nu = \frac{dx}{dt}                or\: \: \int_{0}^{x}dx= \int_{0}^{t}\nu dt

or\: \: x= \int_{0}^{t}\left ( \nu _{0} +gt+ft^{2}\right )dt

x= \nu _{0}t+\frac{gt^{2}}{2}+\frac{ft^{3}}{3}+C

Where  C is the constant of integration

Given : x = 0,t = 0.    C = 0

x= \nu _{0}t+\frac{gt^{2}}{2}+\frac{ft^{3}}{3}

At t = 1 sec

\therefore x= \nu _{0}+\frac{g}{2}+\frac{f}{3}

 

Posted by

Suraj Bhandari

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