Question

The velocity of a particle is $v=v_0+gt+f{t}^2$. If its position is $x=0$ at $t=0$, then its displacement after unit time $(t=1)$ is

• A. $v_0+g/2+f$
• B. $v_0+2g+3f$
• C. $v_0+g/2+f/3$
• D. $v_0+g/2+2f$

Answer 1

The correct option is C $v_0+g/2+f/3$

We know that, $v=\frac{dx}{dt}$

$\Rightarrow dx=vdt$
$\text{Integrating,}{\int }_0^{x}dx={\int }_0^{t}vdt$

$\Rightarrow x={\int }_0^t(v_0+gt+f{t}^2)dt={[v_{0}t+\frac{g{t}^2}{2}+\frac{f{t}^3}{3}]}_0^t$

$\Rightarrow x=v_{0}t+\frac{g{t}^2}{2}+\frac{f{t}^3}{3}$

$\text{At}t=1,$

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

Hence, $\left(C\right)$ is the correct answer.