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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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