Question

At $t=0$, a particle starts from $(1,0)$ and moves towards positive $x-$axis with speed of $v=3t^2+2t\text{m/s}$. The final position of the particle as a function of time is

• A. $\frac{3t^3}{2}-2t^2+1\text{m}$
• B. $\frac{3t^3}{2}+2t^2+1\text{m}$
• C. $t^3+t^2+1\text{m}$
• D. $t^3-t^2+1\text{m}$

Answer 1

The correct option is C $t^3+t^2+1\text{m}$

Given, particle starts from $(1,0)$, so the initial position of the particle is $x_i=1$
As we know that,
$\int dx={\int }_0^{t}v\left(t\right)dt$
$\Rightarrow {\int }_{x_i}^{x_f}dx={\int }_0^t(3t^2+2t)dt$
$\Rightarrow (x_f-x_i)=[t^3+t^2{]}_0^t$
$\Rightarrow x_f-1=t^3+t^2\Rightarrow x_f=t^3+t^2+1\text{m}$