Question

The velocity of a particle is $v=v_0+ct+g{t}^2$. If its position is $x=0$ at $t=0$, then its displacement at time $(t=2\text{sec})$ is

• A. $2v_0+2c+\frac{8g}{3}$
• B. $2v_0+c+\frac{g}{3}$
• C. $2v_0+2c+\frac{2g}{3}$
• D. $2v_0+2c+\frac{6g}{3}$

Answer 1

The correct option is A $2v_0+2c+\frac{8g}{3}$

As we know that the displacement of the particle is given by
$x\left(t\right)={\int }_0^t\left(v\right(t\left)\right)dt\Rightarrow {\int }_0^t(v_0+ct+g{t}^2)dt$
$\Rightarrow x\left(t\right)=v_{0}t+\frac{c{t}^2}{2}+\frac{g{t}^3}{3}$
$\therefore$ Displacement of particle after time $t=2\text{sec}$ is given by
$x{|}_{t=2sec}=2v_0+2c+\frac{8g}{3}$