Velocity of a particle is v=v_0 +gt+ft^2 . If its position is x= 0 at t= 0 , then its displac

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2.Motion in Straight Line

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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

$v_0$$+$$\frac{g}{2}$$+f$

$v_0 +2g+3f$

$v_0 $$+$$\frac{g}{2}$+$\frac{f}{3}$

$v_0 +g+f$

(AIEEE-2007)

Solution

$\begin{array}{l}
We\,know\,that,\,v = \frac{{dx}}{{dt}} \Rightarrow dx = v\,dt\\
Integrating\,\int\limits_0^x {dx = \int\limits_0^t {v\,dt} } \\
or\,\,\,\,\,x = \int\limits_0^t {\left( {{v_0} + gt + f{t^2}} \right)dt = \left[ {{v_0}t + \frac{{g{t^2}}}{2} + \frac{{f{t^3}}}{3}} \right]_0^3} \\
or,\,\,\,\,x = {v_0}t + \frac{{g{t^2}}}{2} + \frac{{f{t^3}}}{3}\\
At\,t = 1,\,x = {v_0} + \frac{g}{2} + \frac{f}{3}.
\end{array}$

Standard 11

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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

Solution

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