Question

At $t=0$, a particle starts from $(-2,0)$ and moves towards positive $x-$axis with speed of $v=6t^2+4t\text{m/s}$. The final position of the particle and the distance travelled by the particle respectively, as a function of time are

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

Answer 1

The correct option is B $2t^3+2t^2-2\text{m},2t^3+2t^2\text{m}$

Given, particle starts from $(-2,0)$, so the initial position of the particle is $x_i=-2$
As we know that,
$\int dx={\int }_0^{t}v\left(t\right)dt$
$\Rightarrow {\int }_{x_i}^{x_f}dx={\int }_0^t(6t^2+4t)dt$
$\Rightarrow (x_f-x_i)=[2t^3+2t^2{]}_0^t$
$\Rightarrow x_f+2=2t^3+2t^2\Rightarrow x_f=2t^3+2t^2-2\text{m}$
And, the distance travelled by the particle is given by
${\int }_0^{t}v\left(t\right)dt={\int }_0^t(6t^2+4t)dt=[2t^3+2t^2{]}_0^t=2t^3+2t^2\text{m}$